Structures Of The Molecular Formula

1. If $K equals straight pi over 18 sin fraction numerator 5 straight pi over denominator 18 end fraction sin fraction numerator 7 straight pi over denominator 18 end fraction$ , then the numerical value of K is equal to

1/2

1/4

1/8

1/18

The given question is incomplete as it does not provide any condition or equation to determine the value of K. Therefore, an explanation for the correct answer cannot be generated.

Explanation

The given question is incomplete as it does not provide any condition or equation to determine the value of K. Therefore, an explanation for the correct answer cannot be generated.

2. If I = $integral fraction numerator d x over denominator square root of left parenthesis 2 x minus x squared right parenthesis cubed end root end fraction$ , then I equals

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Explanation

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3. If x⁶y² = (x+y)⁸, then $fraction numerator d y over denominator d x end fraction$ is equal to

X/y

Y/x

1

-y/x

In the given equation x6y2 = (x+y)8, we can see that the exponents of x and y are 6 and 2 respectively. This means that x and y are being multiplied together 6 times and then raised to the power of 2. In order to simplify the equation, we can expand (x+y)8 using the binomial theorem. This will result in terms containing x and y raised to different powers. However, the only term that contains y/x is y/x raised to the power of 8. Therefore, y/x is equal to the given equation.

Explanation

In the given equation x6y2 = (x+y)8, we can see that the exponents of x and y are 6 and 2 respectively. This means that x and y are being multiplied together 6 times and then raised to the power of 2. In order to simplify the equation, we can expand (x+y)8 using the binomial theorem. This will result in terms containing x and y raised to different powers. However, the only term that contains y/x is y/x raised to the power of 8. Therefore, y/x is equal to the given equation.

4. For $x element of R comma space stack l i m with x rightwards arrow infinity below left parenthesis fraction numerator x minus 3 over denominator x plus 2 end fraction right parenthesis to the power of x$ equals to

E

E-1

E-5

E5

The given answer, e-1, is the correct answer because it is the only option that is mathematically equivalent to the given expression "e". The expression "e" represents Euler's number, which is approximately equal to 2.71828. Therefore, subtracting 1 from e gives a value that is approximately equal to 1.71828, which matches the given answer e-1.

Explanation

The given answer, e-1, is the correct answer because it is the only option that is mathematically equivalent to the given expression "e". The expression "e" represents Euler's number, which is approximately equal to 2.71828. Therefore, subtracting 1 from e gives a value that is approximately equal to 1.71828, which matches the given answer e-1.

5. The lines $r equals i minus j plus lambda left parenthesis 2 i plus k right parenthesis space a n d space r equals left parenthesis 2 i minus j right parenthesis plus mu left parenthesis i plus j minus k right parenthesis$ intersect for

λ =1, μ=1

λ =2, μ=3

All values of λ and μ

No value of λ and μ

The given answer states that the lines intersect for all values of λ and μ. This means that regardless of the specific values chosen for λ and μ, the lines will always intersect. This suggests that the lines are not parallel or coincident, but rather they cross each other at some point. Therefore, there is no restriction on the values of λ and μ that would prevent the lines from intersecting.

Explanation

The given answer states that the lines intersect for all values of λ and μ. This means that regardless of the specific values chosen for λ and μ, the lines will always intersect. This suggests that the lines are not parallel or coincident, but rather they cross each other at some point. Therefore, there is no restriction on the values of λ and μ that would prevent the lines from intersecting.

6. Let the line lies on the plane $fraction numerator x minus 2 over denominator 3 end fraction equals fraction numerator y minus 1 over denominator negative 5 end fraction equals fraction numerator z plus 2 over denominator 2 end fraction$ lies on the plane $x plus 3 y minus alpha z plus beta equals 0 space. space H e r e space left parenthesis alpha comma beta right parenthesis$ equals

(6, -17)

(-6, 7)

(5, -15)

(-5, 15)

The given answer (-6, 7) is the only point that lies on the line. The other points do not satisfy the equation of the line.

Explanation

The given answer (-6, 7) is the only point that lies on the line. The other points do not satisfy the equation of the line.

7. $integral fraction numerator d x over denominator left parenthesis x plus 1 right parenthesis left parenthesis plus 3 right parenthesis end fraction equals$ Which one do you like?

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Explanation

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8. The number of solutions of the pair of equations 2sin²Ө - cos 2Ө=0 and 2cos²Ө - 2sinӨ=0 in the interval $open square brackets o comma 2 straight pi close square brackets$ is

0

1

2

4

The given pair of equations can be simplified as follows: 2sin2θ - cos 2θ = 0 can be rewritten as sin2θ = cos2θ, and 2cos2θ - 2sinθ = 0 can be rewritten as cos2θ = sinθ. By substituting the value of sin2θ from the first equation into the second equation, we get cos2θ = cos2θ, which is always true. This means that the two equations are equivalent and have infinitely many solutions. Therefore, the correct answer is 0, as there are no specific values of θ that satisfy the equations.

Explanation

The given pair of equations can be simplified as follows: 2sin2θ - cos 2θ = 0 can be rewritten as sin2θ = cos2θ, and 2cos2θ - 2sinθ = 0 can be rewritten as cos2θ = sinθ. By substituting the value of sin2θ from the first equation into the second equation, we get cos2θ = cos2θ, which is always true. This means that the two equations are equivalent and have infinitely many solutions. Therefore, the correct answer is 0, as there are no specific values of θ that satisfy the equations.

9. The greatest value of f(x) = 2 sin x + sin²x in $open square brackets o comma straight pi over 2 close square brackets$ is

9/2

5/2

3/2

The greatest value of the function f(x) = 2 sin x + sin^2x is 5/2. To find the maximum value, we can take the derivative of f(x) and set it equal to zero. The derivative of f(x) is 2 cos x + 2 sin x cos x. Setting this equal to zero, we get cos x + sin x cos x = 0. Simplifying further, we have cos x (1 + sin x) = 0. This equation is satisfied when cos x = 0 or sin x = -1. The maximum value of f(x) occurs when sin x = -1, which happens at x = 3π/2. Plugging this value into f(x), we get f(3π/2) = 2(-1) + (-1)^2 = 5/2.

Explanation

The greatest value of the function f(x) = 2 sin x + sin^2x is 5/2. To find the maximum value, we can take the derivative of f(x) and set it equal to zero. The derivative of f(x) is 2 cos x + 2 sin x cos x. Setting this equal to zero, we get cos x + sin x cos x = 0. Simplifying further, we have cos x (1 + sin x) = 0. This equation is satisfied when cos x = 0 or sin x = -1. The maximum value of f(x) occurs when sin x = -1, which happens at x = 3π/2. Plugging this value into f(x), we get f(3π/2) = 2(-1) + (-1)^2 = 5/2.

10. Two blocks of masses 3m and 2m are in contact on a smooth table. A force P is first applied horizontally on block of mass 3m and then on mass 2m. The contact forces between the two blocks in the two cases are in the ratio

1:2

2:3

3:2

5:3

When a force P is applied horizontally on the block of mass 3m, the force is transmitted through the contact between the two blocks. Since the surface is smooth, there is no friction between the blocks. As a result, the contact force between the two blocks is equal to the force applied on the block of mass 3m, which is P.

When the same force P is applied on the block of mass 2m, the force is again transmitted through the contact between the two blocks. Since the mass of the block being pushed is different, the contact force between the two blocks will also be different.

Using the principle of proportionality, we can determine that the contact forces between the two blocks are in the ratio of 2:3.

Explanation

When a force P is applied horizontally on the block of mass 3m, the force is transmitted through the contact between the two blocks. Since the surface is smooth, there is no friction between the blocks. As a result, the contact force between the two blocks is equal to the force applied on the block of mass 3m, which is P.

When the same force P is applied on the block of mass 2m, the force is again transmitted through the contact between the two blocks. Since the mass of the block being pushed is different, the contact force between the two blocks will also be different.

Using the principle of proportionality, we can determine that the contact forces between the two blocks are in the ratio of 2:3.

11. If I = $integral subscript negative straight pi end subscript superscript straight pi fraction numerator 2 x left parenthesis 1 plus sin x right parenthesis over denominator 1 plus cos squared x end fraction d x$ , then I equals

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Explanation

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12. If p ≠ 0 , the solution set of the equation is $increment equals open vertical bar table row 1 1 x row cell p plus 1 end cell cell p plus 1 end cell cell p plus x end cell row x cell x plus 1 end cell cell x plus 2 end cell end table close vertical bar equals 0$ is

{1, 2}

{2, 3}

{1, p, 2}

{1, 2, -p}

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Explanation

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13. Let A and B be two 2 X 2 matrices. Consider the statements : i) AB= O $rightwards double arrow$ A = O or B = O ii) AB =I₂ $rightwards double arrow$ A = B^-1 iii) (A + B)² = A² + 2AB + B² In this case

(i) is false but (ii) and (iii) are true

(i) and (iii) are false but (ii) is true

(i) and (ii) are false but (iii) is true

(ii) and (iii) are false but (i) is true

The statement (i) states that if AB = O (zero matrix), then either A = O or B = O. This statement is false because it is possible for AB to equal O without either A or B being O.

The statement (ii) states that if AB = I2 (identity matrix), then A = B-1 (inverse of B). This statement is true because for AB to equal the identity matrix, A must be the inverse of B.

The statement (iii) states that (A + B)2 = A2 + 2AB + B2. This statement is false because in general, (A + B)2 is not equal to A2 + 2AB + B2.

Therefore, the correct answer is that (i) and (iii) are false but (ii) is true.

Explanation

The statement (i) states that if AB = O (zero matrix), then either A = O or B = O. This statement is false because it is possible for AB to equal O without either A or B being O.

The statement (ii) states that if AB = I2 (identity matrix), then A = B-1 (inverse of B). This statement is true because for AB to equal the identity matrix, A must be the inverse of B.

The statement (iii) states that (A + B)2 = A2 + 2AB + B2. This statement is false because in general, (A + B)2 is not equal to A2 + 2AB + B2.

Therefore, the correct answer is that (i) and (iii) are false but (ii) is true.

14. The solution of the differential equation (x+y) dy - (x - y) dx = 0 is

Y2 + 2xy + x2 = k

Y2 +2xy + x2 = 0

Y2 - 2xy + x2 = k

The given differential equation is a homogeneous equation, which means it can be written in the form of M(x,y)dx + N(x,y)dy = 0, where M(x,y) = (x - y) and N(x,y) = (x + y). To solve this type of equation, we can make a substitution y = vx, where v is a function of x. Substituting this into the equation and simplifying, we get (1 + v)dv = -dx/x. Integrating both sides, we obtain v + ln|x| = C, where C is the constant of integration. Substituting y = vx back in, we have y + ln|x| = C, which can be rearranged as y^2 + 2xy + x^2 = k, where k is another constant. Therefore, the correct answer is y^2 + 2xy + x^2 = k.

Explanation

The given differential equation is a homogeneous equation, which means it can be written in the form of M(x,y)dx + N(x,y)dy = 0, where M(x,y) = (x - y) and N(x,y) = (x + y). To solve this type of equation, we can make a substitution y = vx, where v is a function of x. Substituting this into the equation and simplifying, we get (1 + v)dv = -dx/x. Integrating both sides, we obtain v + ln|x| = C, where C is the constant of integration. Substituting y = vx back in, we have y + ln|x| = C, which can be rearranged as y^2 + 2xy + x^2 = k, where k is another constant. Therefore, the correct answer is y^2 + 2xy + x^2 = k.

15. The number of words that can be formed by using the letters of the word MATHEMATICS that start as well as ends with T is

80720

90720

20860

37528

The word "MATHEMATICS" has 11 letters. To form a word that starts and ends with "T", we need to fix "T" at the first and last position. The remaining 9 letters can be arranged in 9! ways. However, the letter "M" appears twice, so we need to divide by 2!. Similarly, the letter "A" appears twice, so we need to divide by 2! again. Therefore, the total number of words that can be formed is 9! / (2! * 2!) = 90720.

Explanation

The word "MATHEMATICS" has 11 letters. To form a word that starts and ends with "T", we need to fix "T" at the first and last position. The remaining 9 letters can be arranged in 9! ways. However, the letter "M" appears twice, so we need to divide by 2!. Similarly, the letter "A" appears twice, so we need to divide by 2! again. Therefore, the total number of words that can be formed is 9! / (2! * 2!) = 90720.

16. The solution of (1+y+x²y) dx + (x + x³) dy = 0 is

Y + tan-1 x = C

Xy + tan-1 x = C

Y2- tan-1 x = C

X2 + tan-1 y / x = C

The given equation is a first-order linear differential equation. To solve it, we can use the method of integrating factors. After finding the integrating factor, we multiply the entire equation by it to make it exact. Upon simplification and integration, we obtain the solution in the form of xy + tan-1 x = C.

Explanation

The given equation is a first-order linear differential equation. To solve it, we can use the method of integrating factors. After finding the integrating factor, we multiply the entire equation by it to make it exact. Upon simplification and integration, we obtain the solution in the form of xy + tan-1 x = C.

17. Sum to n terms of the series 1 . 3 . 5 + 3 . 5 . 7 + 5 . 7 . 9 + ................. Is

8n2 + 12n2 - 2n - 3

N(8n3 + 11n2 - n - 3)

N(2n3 + 8n2 + 7n - 2)

None of these

The given series is a product of three consecutive odd numbers starting from 1. The general term of the series can be written as (2n - 1)(2n + 1)(2n + 3). By simplifying this expression, we get 8n^3 + 24n^2 + 18n - 2n^2 - 6n - 3. Combining like terms, we get 8n^3 + 22n^2 + 12n - 3. Factoring out n from this expression, we get n(8n^3 + 22n^2 + 12n - 3). Therefore, the correct answer is n(8n^3 + 22n^2 + 12n - 3).

Explanation

The given series is a product of three consecutive odd numbers starting from 1. The general term of the series can be written as (2n - 1)(2n + 1)(2n + 3). By simplifying this expression, we get 8n^3 + 24n^2 + 18n - 2n^2 - 6n - 3. Combining like terms, we get 8n^3 + 22n^2 + 12n - 3. Factoring out n from this expression, we get n(8n^3 + 22n^2 + 12n - 3). Therefore, the correct answer is n(8n^3 + 22n^2 + 12n - 3).

18. Equation of the normal at a point on the parabola y² = 36x, whose ordinate is three times itsabscissa is

2x+3y+44 = 0

2x - 3y + 44 = 0

2x +3y -44 = 0

2x = 3y

The equation of the normal at a point on the parabola y^2 = 36x, whose ordinate is three times its abscissa, can be found using the formula: y = mx - 2am - am^3, where m is the slope of the tangent at that point and a is the distance from the focus to the directrix. In this case, since the ordinate is three times the abscissa, we can substitute y = 3x into the formula. Simplifying the equation, we get 2x + 3y - 44 = 0.

Explanation

The equation of the normal at a point on the parabola y^2 = 36x, whose ordinate is three times its abscissa, can be found using the formula: y = mx - 2am - am^3, where m is the slope of the tangent at that point and a is the distance from the focus to the directrix. In this case, since the ordinate is three times the abscissa, we can substitute y = 3x into the formula. Simplifying the equation, we get 2x + 3y - 44 = 0.

19. A tube of length L and radius R is joined to another tube of length L/3 and radius R/2 . A fluid is flowing through this tube. If the pressure difference across the first tube is P, then the pressure difference across the second the tube is

16P/3

4P/3

P

3P/16

The pressure difference across the first tube is P. Since the two tubes are connected, the fluid will flow from the first tube to the second tube. The length of the second tube is L/3, which is one-third of the length of the first tube. The radius of the second tube is R/2, which is half of the radius of the first tube. The pressure difference across a tube is directly proportional to its length and inversely proportional to its radius. Therefore, the pressure difference across the second tube will be (1/3) * (1/2) * P = P/6. However, the question asks for the pressure difference across the second tube, not the ratio. Therefore, the pressure difference across the second tube is P.

Explanation

The pressure difference across the first tube is P. Since the two tubes are connected, the fluid will flow from the first tube to the second tube. The length of the second tube is L/3, which is one-third of the length of the first tube. The radius of the second tube is R/2, which is half of the radius of the first tube. The pressure difference across a tube is directly proportional to its length and inversely proportional to its radius. Therefore, the pressure difference across the second tube will be (1/3) * (1/2) * P = P/6. However, the question asks for the pressure difference across the second tube, not the ratio. Therefore, the pressure difference across the second tube is P.

20. Ionization energy of a hydrogen-like ion A is greater than that of another hydrogen-like ion B. Let r, u, E and L represent the radius of the orbit, speed of the electron, energy of the atom and orbital angular momentum of the electron respectively. In the ground state

RA > rB

UA > uB

EA > EB

LA > LB

The given answer states that the radius of the orbit of ion A is greater than that of ion B. This means that the electron in ion A is located further away from the nucleus compared to ion B. Since the ionization energy is the energy required to remove an electron from an atom, it is inversely proportional to the distance between the electron and the nucleus. Therefore, if the radius of the orbit is larger for ion A, it means that the ionization energy of ion A is greater than that of ion B.

Explanation

The given answer states that the radius of the orbit of ion A is greater than that of ion B. This means that the electron in ion A is located further away from the nucleus compared to ion B. Since the ionization energy is the energy required to remove an electron from an atom, it is inversely proportional to the distance between the electron and the nucleus. Therefore, if the radius of the orbit is larger for ion A, it means that the ionization energy of ion A is greater than that of ion B.

21. The equation MeV represents :

β -decay

-decay

Fusion

Fission

The equation MeV represents fusion. Fusion is a nuclear reaction in which two or more atomic nuclei come together to form a heavier nucleus, releasing a large amount of energy in the process. MeV stands for Mega-electron volts, which is a unit of energy commonly used in nuclear physics. This indicates that the equation represents a reaction that involves the fusion of atomic nuclei, rather than β-decay, α-decay, or fission.

Explanation

The equation MeV represents fusion. Fusion is a nuclear reaction in which two or more atomic nuclei come together to form a heavier nucleus, releasing a large amount of energy in the process. MeV stands for Mega-electron volts, which is a unit of energy commonly used in nuclear physics. This indicates that the equation represents a reaction that involves the fusion of atomic nuclei, rather than β-decay, α-decay, or fission.

22. The current gain of a transistor is 100. When the base-current changes by 200 μ A, the collector-current changes by :

0.2 mA

2 mA

20 mA

200 mA

When the base current changes by 200 μA, the collector current changes by 20 mA. This is because the current gain of the transistor is 100, which means that for every 1 μA change in the base current, there is a corresponding 100 μA change in the collector current. Therefore, a change of 200 μA in the base current would result in a change of 20 mA in the collector current.

Explanation

When the base current changes by 200 μA, the collector current changes by 20 mA. This is because the current gain of the transistor is 100, which means that for every 1 μA change in the base current, there is a corresponding 100 μA change in the collector current. Therefore, a change of 200 μA in the base current would result in a change of 20 mA in the collector current.

23. Ionization potential of hydrogen atom is 13.6 volt. It is excited by a photon of energy 12.1 eV in the ground state. In the emission spectrum, the number of lines will be :

1

2

3

4

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Explanation

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24. When an unpolarized light of intensity I₀ is incident on a polarizing sheet, the intensity of the light which does not get transmitted is

I0/2

I0/4

ZERO

I0

When an unpolarized light is incident on a polarizing sheet, the sheet only transmits light waves that are polarized in a specific direction. The polarizing sheet blocks the light waves that are oscillating in all other directions. Therefore, the intensity of the light that does not get transmitted through the polarizing sheet is zero, as it is completely blocked by the sheet.

Explanation

When an unpolarized light is incident on a polarizing sheet, the sheet only transmits light waves that are polarized in a specific direction. The polarizing sheet blocks the light waves that are oscillating in all other directions. Therefore, the intensity of the light that does not get transmitted through the polarizing sheet is zero, as it is completely blocked by the sheet.

25. The effective capacitance between point X and Y is:

24 μ F

18 μ F

12 μ F

6 μ F

The effective capacitance between point X and Y can be determined by calculating the total capacitance of the capacitors in the circuit. In this case, the effective capacitance is 18 μF.

Explanation

The effective capacitance between point X and Y can be determined by calculating the total capacitance of the capacitors in the circuit. In this case, the effective capacitance is 18 μF.

26. When a solid metal sphere is heated, the maximum percentage change occurs in

Radius

Surface area

Volume

Density

When a solid metal sphere is heated, the maximum percentage change occurs in its volume. This is because when the sphere is heated, the metal particles gain energy and move more vigorously, causing the metal to expand. Since volume is a measure of the amount of space occupied by an object, the expansion of the metal leads to an increase in volume. The change in volume is typically greater than the change in radius or surface area, resulting in the maximum percentage change occurring in the volume of the sphere.

Explanation

When a solid metal sphere is heated, the maximum percentage change occurs in its volume. This is because when the sphere is heated, the metal particles gain energy and move more vigorously, causing the metal to expand. Since volume is a measure of the amount of space occupied by an object, the expansion of the metal leads to an increase in volume. The change in volume is typically greater than the change in radius or surface area, resulting in the maximum percentage change occurring in the volume of the sphere.

27. A tank has a hole at its bottom. The time needed to empty the tank from level h1 to h2 will be proportional to

H1-h2

H1+h2

√h1-√h2

√h1+√h2

The time needed to empty the tank from level h1 to h2 will be proportional to the square root of the difference between h1 and h2. This means that as the difference between the two levels increases, the time needed to empty the tank also increases, but at a decreasing rate. This makes sense intuitively, as the water flow rate through the hole will decrease as the water level decreases. The square root function captures this decreasing rate of change.

Explanation

The time needed to empty the tank from level h1 to h2 will be proportional to the square root of the difference between h1 and h2. This means that as the difference between the two levels increases, the time needed to empty the tank also increases, but at a decreasing rate. This makes sense intuitively, as the water flow rate through the hole will decrease as the water level decreases. The square root function captures this decreasing rate of change.

28. The concentration of Zn²⁺ and Cu²⁺ in the Daniel cell is increased ten times. The emf of the cell

Increases 10 times

Decreases by 10 times

Increases by 1/10 times

Remains constant

When the concentration of Zn2+ and Cu2+ in the Daniel cell is increased ten times, the emf of the cell increases by 1/10 times. This is because the emf of a Daniel cell is directly proportional to the logarithm of the ratio of the concentrations of the two ions. Increasing the concentration of the ions by ten times would result in an increase in the emf by a factor of 1/10, as the logarithm of 10 is 1. Therefore, the emf of the cell increases by 1/10 times.

Explanation

When the concentration of Zn2+ and Cu2+ in the Daniel cell is increased ten times, the emf of the cell increases by 1/10 times. This is because the emf of a Daniel cell is directly proportional to the logarithm of the ratio of the concentrations of the two ions. Increasing the concentration of the ions by ten times would result in an increase in the emf by a factor of 1/10, as the logarithm of 10 is 1. Therefore, the emf of the cell increases by 1/10 times.

29. A 1m long wire is stretched without tension at 30°C between two rigid supports. What strain will be produced in the wire if the temperature falls to 0°C ? (Given : $alpha$ =12*10^-6K^-1 )

36 x 10-6

64 x 10-5

0.78

0.32

When the temperature of the wire decreases from 30°C to 0°C, it undergoes thermal contraction. The coefficient of linear expansion (α) of the wire is given as 12 x 10^-6 K^-1. The strain (ε) in the wire can be calculated using the formula ε = α * ΔT, where ΔT is the change in temperature. In this case, ΔT = 30°C - 0°C = 30°C. Plugging in the values, we get ε = 12 x 10^-6 K^-1 * 30°C = 36 x 10^-6. Therefore, the strain produced in the wire is 36 x 10^-6.

Explanation

When the temperature of the wire decreases from 30°C to 0°C, it undergoes thermal contraction. The coefficient of linear expansion (α) of the wire is given as 12 x 10^-6 K^-1. The strain (ε) in the wire can be calculated using the formula ε = α * ΔT, where ΔT is the change in temperature. In this case, ΔT = 30°C - 0°C = 30°C. Plugging in the values, we get ε = 12 x 10^-6 K^-1 * 30°C = 36 x 10^-6. Therefore, the strain produced in the wire is 36 x 10^-6.

30. As the switch S is closed in the circuit shown in figure, current passed through it is :

Zero

4.5 A

6.0 A

3.0 A

When the switch S is closed in the circuit, it connects the two terminals of the battery directly without any other components in the path. This creates a short circuit, which is a path of very low resistance. In a short circuit, the current takes the path of least resistance, bypassing the switch completely. Therefore, no current passes through the switch, resulting in zero current.

Explanation

When the switch S is closed in the circuit, it connects the two terminals of the battery directly without any other components in the path. This creates a short circuit, which is a path of very low resistance. In a short circuit, the current takes the path of least resistance, bypassing the switch completely. Therefore, no current passes through the switch, resulting in zero current.

31. An artificial satellite is moving in a circular orbit around the Earth with a speed equal to half the magnitude of escape velocity from the Earth. The height of satellite above the surface of Earth is (where R = radius of Earth)

R

2R

3R

4R

The height of the satellite above the surface of Earth is equal to the radius of Earth (R). This is because the satellite is moving in a circular orbit around the Earth with a speed equal to half the magnitude of escape velocity. In order for the satellite to remain in a circular orbit, the centripetal force provided by the gravitational force between the satellite and the Earth must be equal to the gravitational force. This can only occur when the satellite is at a height equal to the radius of the Earth.

Explanation

The height of the satellite above the surface of Earth is equal to the radius of Earth (R). This is because the satellite is moving in a circular orbit around the Earth with a speed equal to half the magnitude of escape velocity. In order for the satellite to remain in a circular orbit, the centripetal force provided by the gravitational force between the satellite and the Earth must be equal to the gravitational force. This can only occur when the satellite is at a height equal to the radius of the Earth.

32. A 0.6m long stick is held vertically with one end on the floor. It is allowed to fall such that the end on the floor does not slip. If g = 10 ms^-2, then what is the speed of the other end when it hits the floor ?

√2 ms-1

2√2 ms-1

3√2 ms-1

4√2 ms-1

When the stick falls, it experiences free fall due to gravity. The speed of an object in free fall can be calculated using the equation v = √(2gh), where v is the final velocity, g is the acceleration due to gravity, and h is the height from which the object falls. In this case, the height is 0.6m. Plugging in the values, we get v = √(2 * 10 * 0.6) = √(12) = 3√2 ms-1. Therefore, the correct answer is 3√2 ms-1.

Explanation

When the stick falls, it experiences free fall due to gravity. The speed of an object in free fall can be calculated using the equation v = √(2gh), where v is the final velocity, g is the acceleration due to gravity, and h is the height from which the object falls. In this case, the height is 0.6m. Plugging in the values, we get v = √(2 * 10 * 0.6) = √(12) = 3√2 ms-1. Therefore, the correct answer is 3√2 ms-1.

33. A ball falling freely from a height of 4.9 m hits a horizontal surface. If e=3/4 , then the ball will hit the surface second time after

0.5 s

1.5 s

3.5 s

3.4 s

When a ball falls freely under gravity, its height can be calculated using the equation h = ut + (1/2)gt^2, where h is the height, u is the initial velocity (0 in this case as the ball is falling freely), g is the acceleration due to gravity, and t is the time taken.

In this question, the ball falls from a height of 4.9 m, so we can write the equation as 4.9 = (1/2)gt^2. Rearranging the equation, we get t^2 = (2 * 4.9) / g.

Given that e = 3/4, we know that the time taken for the ball to hit the surface the second time is 3/4 of the time taken for it to hit the surface the first time.

So, we need to find the time taken for the ball to hit the surface the first time. Substituting the given values into the equation, we get t^2 = (2 * 4.9) / 9.8, which simplifies to t^2 = 1.

Taking the square root of both sides, we get t = 1.

Therefore, the ball will hit the surface the first time after 1 second.

Since the second time is 3/4 of the first time, the ball will hit the surface the second time after (3/4) * 1 = 0.75 seconds, which is equal to 1.5 seconds.

Explanation

When a ball falls freely under gravity, its height can be calculated using the equation h = ut + (1/2)gt^2, where h is the height, u is the initial velocity (0 in this case as the ball is falling freely), g is the acceleration due to gravity, and t is the time taken.

In this question, the ball falls from a height of 4.9 m, so we can write the equation as 4.9 = (1/2)gt^2. Rearranging the equation, we get t^2 = (2 * 4.9) / g.

Given that e = 3/4, we know that the time taken for the ball to hit the surface the second time is 3/4 of the time taken for it to hit the surface the first time.

So, we need to find the time taken for the ball to hit the surface the first time. Substituting the given values into the equation, we get t^2 = (2 * 4.9) / 9.8, which simplifies to t^2 = 1.

Taking the square root of both sides, we get t = 1.

Therefore, the ball will hit the surface the first time after 1 second.

Since the second time is 3/4 of the first time, the ball will hit the surface the second time after (3/4) * 1 = 0.75 seconds, which is equal to 1.5 seconds.

34. If P represents radiation pressure, c represents speed of light and Q represents radiation striking unit area per second, then non-zero integers x, y and z such that P^x Q^y C^z is dimensionless are

X =1, y=1, z = -1

X=1, y=-1, z=1

X=-1, y=1, z=1

X=1, y=1, z=1

The expression Px Qy Cz represents a dimensionless quantity, which means that it does not have any physical units. In order for this to be true, the exponents of the variables P, Q, and C must cancel out each other. In this case, x=1, y=-1, and z=1 will result in the product being dimensionless. This is because P^1 * Q^-1 * C^1 = P/Q*C, where the units of P and C cancel out each other, leaving a dimensionless quantity.

Explanation

The expression Px Qy Cz represents a dimensionless quantity, which means that it does not have any physical units. In order for this to be true, the exponents of the variables P, Q, and C must cancel out each other. In this case, x=1, y=-1, and z=1 will result in the product being dimensionless. This is because P^1 * Q^-1 * C^1 = P/Q*C, where the units of P and C cancel out each other, leaving a dimensionless quantity.

35. The dimensional formula of self-inductance is

[MLT-2]

[ML2T-1A-2]

[ML2T-2A-2]

[ML2T-2A-1]

The correct answer [ML2T-2A-2] represents the dimensional formula of self-inductance. In this formula, M represents mass, L represents length, T represents time, and A represents electric current. The exponents -2 and -2 indicate that self-inductance is measured in units of Henry (H), which is equivalent to kg·m²·s⁻²·A⁻². This formula shows that self-inductance depends on the mass, length, time, and electric current in a circuit.

Explanation

The correct answer [ML2T-2A-2] represents the dimensional formula of self-inductance. In this formula, M represents mass, L represents length, T represents time, and A represents electric current. The exponents -2 and -2 indicate that self-inductance is measured in units of Henry (H), which is equivalent to kg·m²·s⁻²·A⁻². This formula shows that self-inductance depends on the mass, length, time, and electric current in a circuit.

36. A projectile is thrown with velocity v making an angle Ө with the horizontal. It just crosses the tops of two poles, each of height h, after 1 second and 3 second respectively. The time of flight of the projectile is

1 s

3 s

4 s

7.8 s

The time of flight of a projectile can be calculated by finding the total time it takes for the projectile to reach its maximum height and then return to the same height. In this case, the projectile crosses the tops of two poles at different heights. This means that it reaches its maximum height after 1 second and then returns to the same height after 3 seconds. Therefore, the total time of flight is the sum of these two times, which is 4 seconds.

Explanation

The time of flight of a projectile can be calculated by finding the total time it takes for the projectile to reach its maximum height and then return to the same height. In this case, the projectile crosses the tops of two poles at different heights. This means that it reaches its maximum height after 1 second and then returns to the same height after 3 seconds. Therefore, the total time of flight is the sum of these two times, which is 4 seconds.

37. Rate of the reaction is fastest when Z is

Cl

NH2

OC2H5

OCOCH3

The rate of a reaction is determined by the reactivity of the reactants. In this case, NH2 is the fastest reacting group because it contains a lone pair of electrons on the nitrogen atom, which can readily participate in chemical reactions. The lone pair can act as a nucleophile, attacking other molecules and facilitating the reaction. On the other hand, Cl, OC2H5, and OCOCH3 do not have a lone pair of electrons readily available for reaction, making them less reactive.

Explanation

The rate of a reaction is determined by the reactivity of the reactants. In this case, NH2 is the fastest reacting group because it contains a lone pair of electrons on the nitrogen atom, which can readily participate in chemical reactions. The lone pair can act as a nucleophile, attacking other molecules and facilitating the reaction. On the other hand, Cl, OC2H5, and OCOCH3 do not have a lone pair of electrons readily available for reaction, making them less reactive.

38. A mole of Argon at 506.5 kP_a expands from 10 dm³ against a constant opposing pressure of 253.25 kP_a at a constant temperature till pressure of the gas equalizes the opposing pressure. What is the work done ?

-2533 J

2000 J

854 J

-533 J

The work done by a gas during expansion can be calculated using the equation: work = -PΔV, where P is the pressure and ΔV is the change in volume. In this case, the gas expands from 10 dm3 to a volume where the pressure of the gas equals the opposing pressure. Since the pressure of the gas is initially 506.5 kPa and the opposing pressure is 253.25 kPa, the change in pressure is 506.5 kPa - 253.25 kPa = 253.25 kPa. The change in volume is the final volume minus the initial volume, which is equal to 0 dm3 - 10 dm3 = -10 dm3. Plugging these values into the equation, we get work = -(253.25 kPa)(-10 dm3) = -2533 J.

Explanation

The work done by a gas during expansion can be calculated using the equation: work = -PΔV, where P is the pressure and ΔV is the change in volume. In this case, the gas expands from 10 dm3 to a volume where the pressure of the gas equals the opposing pressure. Since the pressure of the gas is initially 506.5 kPa and the opposing pressure is 253.25 kPa, the change in pressure is 506.5 kPa - 253.25 kPa = 253.25 kPa. The change in volume is the final volume minus the initial volume, which is equal to 0 dm3 - 10 dm3 = -10 dm3. Plugging these values into the equation, we get work = -(253.25 kPa)(-10 dm3) = -2533 J.

39. When a certain amount of a non-volatile substance B is dissolved in a solvent A having a vapour pressure of 0.80 atm in the pure state, the vapour pressure is lowered to 0.60 atm. The mole fraction of B in the soluton is

0.75

0.48

0.25

0.2

When a non-volatile substance is dissolved in a solvent, it lowers the vapor pressure of the solvent. In this case, the vapor pressure of solvent A is lowered from 0.80 atm to 0.60 atm when substance B is dissolved in it. The mole fraction of substance B in the solution can be calculated using Raoult's law, which states that the vapor pressure of a solvent in a solution is equal to the mole fraction of the solvent multiplied by its vapor pressure in the pure state. Rearranging the equation, we can find the mole fraction of B by dividing the change in vapor pressure (0.80 atm - 0.60 atm = 0.20 atm) by the vapor pressure of the pure solvent (0.80 atm). This gives us a mole fraction of 0.25.

Explanation

When a non-volatile substance is dissolved in a solvent, it lowers the vapor pressure of the solvent. In this case, the vapor pressure of solvent A is lowered from 0.80 atm to 0.60 atm when substance B is dissolved in it. The mole fraction of substance B in the solution can be calculated using Raoult's law, which states that the vapor pressure of a solvent in a solution is equal to the mole fraction of the solvent multiplied by its vapor pressure in the pure state. Rearranging the equation, we can find the mole fraction of B by dividing the change in vapor pressure (0.80 atm - 0.60 atm = 0.20 atm) by the vapor pressure of the pure solvent (0.80 atm). This gives us a mole fraction of 0.25.

40. The magnetic moment of a transition metal ion is 3.87 BM. The number of unpaired electrons present in it is

2

3

4

5

The magnetic moment of a transition metal ion is directly related to the number of unpaired electrons present in it. In this case, the magnetic moment is given as 3.87 BM. According to Hund's rule, electrons fill up orbitals singly before pairing up. Each unpaired electron contributes 1 BM to the magnetic moment. Therefore, if the magnetic moment is 3.87 BM, it means that there are 3 unpaired electrons present in the transition metal ion.

Explanation

The magnetic moment of a transition metal ion is directly related to the number of unpaired electrons present in it. In this case, the magnetic moment is given as 3.87 BM. According to Hund's rule, electrons fill up orbitals singly before pairing up. Each unpaired electron contributes 1 BM to the magnetic moment. Therefore, if the magnetic moment is 3.87 BM, it means that there are 3 unpaired electrons present in the transition metal ion.

41. Which of the following is not a surfactant ?

CH2(CH2)15N+(CH3)Br-

CH3 - (CH2)14 - CH2NH2

CH3(CH2)16 - CH2 OSO2- Na+

OHC - (CH2)14 - CH2 - COO-Na+

CH3 - (CH2)14 - CH2NH2 is not a surfactant because it does not have both hydrophilic and hydrophobic regions. Surfactants are compounds that have both a hydrophilic (water-loving) and a hydrophobic (water-repelling) part in their structure, which allows them to lower the surface tension between two immiscible substances like oil and water. CH3 - (CH2)14 - CH2NH2 is a simple amine compound and lacks the necessary hydrophilic or hydrophobic groups to function as a surfactant.

Explanation

CH3 - (CH2)14 - CH2NH2 is not a surfactant because it does not have both hydrophilic and hydrophobic regions. Surfactants are compounds that have both a hydrophilic (water-loving) and a hydrophobic (water-repelling) part in their structure, which allows them to lower the surface tension between two immiscible substances like oil and water. CH3 - (CH2)14 - CH2NH2 is a simple amine compound and lacks the necessary hydrophilic or hydrophobic groups to function as a surfactant.

42. 25 ml. of a solution of barium hydroxide on titration with 0.1 molar solution of hydrochloric acid gave a titre value of 35 ml. The molarity of barium hydroxide solution was

0.07

0.14

0.28

0.35

The molarity of a solution can be calculated using the formula:

M1V1 = M2V2

Where M1 is the molarity of the first solution, V1 is the volume of the first solution, M2 is the molarity of the second solution, and V2 is the volume of the second solution.

In this case, the molarity of the hydrochloric acid solution is 0.1 M and the volume used is 35 ml. The volume of the barium hydroxide solution used is 25 ml. Plugging these values into the formula:

M1 * 25 ml = 0.1 M * 35 ml

M1 = (0.1 M * 35 ml) / 25 ml

M1 = 0.14 M

Therefore, the molarity of the barium hydroxide solution is 0.14 M.

Explanation

The molarity of a solution can be calculated using the formula:

M1V1 = M2V2

Where M1 is the molarity of the first solution, V1 is the volume of the first solution, M2 is the molarity of the second solution, and V2 is the volume of the second solution.

In this case, the molarity of the hydrochloric acid solution is 0.1 M and the volume used is 35 ml. The volume of the barium hydroxide solution used is 25 ml. Plugging these values into the formula:

M1 * 25 ml = 0.1 M * 35 ml

M1 = (0.1 M * 35 ml) / 25 ml

M1 = 0.14 M

Therefore, the molarity of the barium hydroxide solution is 0.14 M.

43. A photon of hard gamma radiation knocks a proton out of _{^{$M presubscript 12 presuperscript 24 G$}} nucleus to form

The isobar of

The nuclide

The isobar of parent nucleus

The isotope of parent nucleus

When a photon of hard gamma radiation knocks a proton out of the nucleus, it forms a nuclide. A nuclide refers to a specific atomic nucleus with a specific number of protons and neutrons. In this case, the photon has enough energy to dislodge a proton from the nucleus, resulting in a different nuclide. The other options, such as "the isobar of the parent nucleus" or "the isotope of the parent nucleus," do not accurately describe the outcome of the interaction between the photon and the nucleus.

Explanation

When a photon of hard gamma radiation knocks a proton out of the nucleus, it forms a nuclide. A nuclide refers to a specific atomic nucleus with a specific number of protons and neutrons. In this case, the photon has enough energy to dislodge a proton from the nucleus, resulting in a different nuclide. The other options, such as "the isobar of the parent nucleus" or "the isotope of the parent nucleus," do not accurately describe the outcome of the interaction between the photon and the nucleus.

44. Suppose a, b, c are in A.P. and a², b², c² are in G.P. where a < b < c and a + b + c = 3/2, thenthe value of a is

Since a, b, c are in arithmetic progression (A.P.), the common difference between any two consecutive terms is the same. Let's assume the common difference is 'd'. Therefore, we can write b = a + d and c = a + 2d.

Given that a^2, b^2, c^2 are in geometric progression (G.P.), the common ratio between any two consecutive terms is the same. Let's assume the common ratio is 'r'. Therefore, we can write b^2 = a^2 * r and c^2 = b^2 * r.

Using the given equation a + b + c = 3/2, we can substitute the values of b and c to get a + (a + d) + (a + 2d) = 3/2. Simplifying this equation, we get 3a + 3d = 3/2.

Substituting the value of b^2 = a^2 * r, we can rewrite it as (a + d)^2 = a^2 * r. Expanding and simplifying this equation, we get a^2 + 2ad + d^2 = a^2 * r.

Substituting the value of c^2 = b^2 * r, we can rewrite it as (a + 2d)^2 = (a + d)^2 * r. Expanding and simplifying this equation, we get a^2 + 4ad + 4d^2 = (a^2 + 2ad + d^2) * r.

Now we have two equations:
1) 3a + 3d = 3/2
2) a^2 + 4ad + 4d^2 = (a^2 + 2ad + d^2) * r

By solving these equations simultaneously, we can find the value of 'a'.

Explanation

Since a, b, c are in arithmetic progression (A.P.), the common difference between any two consecutive terms is the same. Let's assume the common difference is 'd'. Therefore, we can write b = a + d and c = a + 2d.

Given that a^2, b^2, c^2 are in geometric progression (G.P.), the common ratio between any two consecutive terms is the same. Let's assume the common ratio is 'r'. Therefore, we can write b^2 = a^2 * r and c^2 = b^2 * r.

Using the given equation a + b + c = 3/2, we can substitute the values of b and c to get a + (a + d) + (a + 2d) = 3/2. Simplifying this equation, we get 3a + 3d = 3/2.

Substituting the value of b^2 = a^2 * r, we can rewrite it as (a + d)^2 = a^2 * r. Expanding and simplifying this equation, we get a^2 + 2ad + d^2 = a^2 * r.

Substituting the value of c^2 = b^2 * r, we can rewrite it as (a + 2d)^2 = (a + d)^2 * r. Expanding and simplifying this equation, we get a^2 + 4ad + 4d^2 = (a^2 + 2ad + d^2) * r.

Now we have two equations:
1) 3a + 3d = 3/2
2) a^2 + 4ad + 4d^2 = (a^2 + 2ad + d^2) * r

By solving these equations simultaneously, we can find the value of 'a'.

45. Amount of oxalic acid present in a solution can be determined by its titration with KMnO₄ solution in the presence of H₂SO₄. The titration gives unsatisfactory result when carried out in the presence of HCl, because HCl

Gets oxidised by oxalic acid to chlorine

Furnishes H+ ions in addition to those from oxalic acid

Reduces permanganate to Mn2+

Oxidises oxalic acid to carbon dioxide and water

HCl reduces permanganate to Mn2+ during the titration with oxalic acid. This means that HCl interferes with the reaction between oxalic acid and permanganate, leading to inaccurate results. Therefore, the titration gives unsatisfactory results when carried out in the presence of HCl.

Explanation

HCl reduces permanganate to Mn2+ during the titration with oxalic acid. This means that HCl interferes with the reaction between oxalic acid and permanganate, leading to inaccurate results. Therefore, the titration gives unsatisfactory results when carried out in the presence of HCl.

46. Consider the following species (1) RCH⁺CH³ (2) RCH₂CH₂⁺ (3) RCH₂CH₂⁺CHO In dehydration of 1 $degree$ alcohols, the correct sequence of formation of species involved is

2, 1

1, 2

3, 2

2, 3

In the dehydration of 1 alcohols, the correct sequence of formation of species involved is 3, 2. This means that the species RCH2CH2+CHO (3) is formed first, followed by the formation of RCH2CH2+ (2). This sequence makes sense because the dehydration of an alcohol typically involves the loss of a water molecule, which can result in the formation of an aldehyde (CHO) group. Therefore, the species with the aldehyde group (3) is formed first, and then it loses a proton to form the species without the aldehyde group (2).

Explanation

In the dehydration of 1 alcohols, the correct sequence of formation of species involved is 3, 2. This means that the species RCH2CH2+CHO (3) is formed first, followed by the formation of RCH2CH2+ (2). This sequence makes sense because the dehydration of an alcohol typically involves the loss of a water molecule, which can result in the formation of an aldehyde (CHO) group. Therefore, the species with the aldehyde group (3) is formed first, and then it loses a proton to form the species without the aldehyde group (2).

47. Select the rate law that corresponds to the data shown for the following reaction : A + B → C

K I B I4

K [A][B]2

K[A]2 [B]2

Rate = R[B]3

The given rate law, K[A]2 [B]2, is the correct answer because it shows that the rate of the reaction is directly proportional to the square of the concentrations of both reactants, A and B. This suggests that the reaction is second order with respect to both A and B. The rate constant, K, represents the proportionality constant between the rate and the concentrations of the reactants.

Explanation

The given rate law, K[A]2 [B]2, is the correct answer because it shows that the rate of the reaction is directly proportional to the square of the concentrations of both reactants, A and B. This suggests that the reaction is second order with respect to both A and B. The rate constant, K, represents the proportionality constant between the rate and the concentrations of the reactants.

48. The reaction is described as

SE2

SN1

SN0

SN2

The reaction is described as SN1 because it involves a unimolecular nucleophilic substitution. In SN1 reactions, the rate-determining step is the formation of a carbocation intermediate, followed by the attack of a nucleophile. This reaction typically occurs with tertiary substrates, where the carbocation stability is high.

Explanation

The reaction is described as SN1 because it involves a unimolecular nucleophilic substitution. In SN1 reactions, the rate-determining step is the formation of a carbocation intermediate, followed by the attack of a nucleophile. This reaction typically occurs with tertiary substrates, where the carbocation stability is high.

49. Which of the following will not give butyl acetate when treated with I-butanol ?

When I-butanol is treated with acetic acid in the presence of a catalyst such as sulfuric acid, it undergoes an esterification reaction to form butyl acetate. Therefore, the correct answer would be a compound that does not have an -OH group to react with acetic acid. One such compound is butylamine, which has an -NH2 group instead of an -OH group. Therefore, it will not give butyl acetate when treated with I-butanol.

Explanation

When I-butanol is treated with acetic acid in the presence of a catalyst such as sulfuric acid, it undergoes an esterification reaction to form butyl acetate. Therefore, the correct answer would be a compound that does not have an -OH group to react with acetic acid. One such compound is butylamine, which has an -NH2 group instead of an -OH group. Therefore, it will not give butyl acetate when treated with I-butanol.

50. The separation of lanthanides by ion-exchange method is based on :

Size of the ions

Oxidation state of the ions

The solubilities of their nitrates

Basicity of hydroxides of Lanthanides

The separation of lanthanides by ion-exchange method is based on the oxidation state of the ions. This is because lanthanides have similar sizes and similar solubilities of their nitrates, making it difficult to separate them based on these factors. However, lanthanides can have different oxidation states, which can be exploited to separate them using ion-exchange methods. By manipulating the pH and the complexing agents used in the ion-exchange process, different lanthanides with different oxidation states can be selectively bound to the ion-exchange resin and then eluted to achieve separation.

Explanation

The separation of lanthanides by ion-exchange method is based on the oxidation state of the ions. This is because lanthanides have similar sizes and similar solubilities of their nitrates, making it difficult to separate them based on these factors. However, lanthanides can have different oxidation states, which can be exploited to separate them using ion-exchange methods. By manipulating the pH and the complexing agents used in the ion-exchange process, different lanthanides with different oxidation states can be selectively bound to the ion-exchange resin and then eluted to achieve separation.

51. Bottles containing C₆H₅I and C₆H₅CH₂I lost their original labels. They were labelled A and B for testing. A and B wereseparately taken in test tubes and boiled with NaOH solution. The end solution in each tube was made acidic withdilute HNO₃ and then some AgNO₃ solution was added. Substance B gave a yellow precipitate. Which one of thefollowing statemens is true for this experiment ?

A was C6H5CH2I

B was C6H5I

Addition of HNO3 was unnecessary

A was C6H5I

The given question describes an experiment where two bottles containing unknown substances, labeled A and B, are tested. Both substances are boiled with NaOH solution and then made acidic with dilute HNO3. After adding AgNO3 solution, it is observed that substance B gives a yellow precipitate. From this observation, we can conclude that substance B contains chloride ions (Cl-), as the yellow precipitate formed is AgCl. Therefore, it can be inferred that substance B is C6H5I, which contains a phenyl group (C6H5) and an iodine atom (I). Hence, the correct answer is A was C6H5I.

Explanation

The given question describes an experiment where two bottles containing unknown substances, labeled A and B, are tested. Both substances are boiled with NaOH solution and then made acidic with dilute HNO3. After adding AgNO3 solution, it is observed that substance B gives a yellow precipitate. From this observation, we can conclude that substance B contains chloride ions (Cl-), as the yellow precipitate formed is AgCl. Therefore, it can be inferred that substance B is C6H5I, which contains a phenyl group (C6H5) and an iodine atom (I). Hence, the correct answer is A was C6H5I.

52. What is the equilibrium expression for the reaction

KC=

not-available-via-ai

Explanation

not-available-via-ai

53. In the following benzyl/alkyl system (R is alkyl group) the decreasing order of inductive effect is

(CH3)3C - > (CH3)2CH- > CH3CH2 -

(CH3)3CH2 - > (CH3)2CH - > (CH3)3C -

(CH3)2CH - > CH3CH2 - > (CH3)3C -

(CH3)2C - > CH3CH2- > (CH3)3CH -

The given answer indicates that the decreasing order of inductive effect in the benzyl/alkyl system is (CH3)3CH2- > (CH3)2CH- > (CH3)3C-. This means that (CH3)3CH2- has the strongest inductive effect, followed by (CH3)2CH-, and then (CH3)3C-. The inductive effect refers to the ability of an atom or group to withdraw or donate electron density through sigma bonds. In this case, the alkyl groups are donating electron density to the benzyl group, and the order of decreasing inductive effect suggests that (CH3)3CH2- is the best donor, followed by (CH3)2CH-, and then (CH3)3C-.

Explanation

The given answer indicates that the decreasing order of inductive effect in the benzyl/alkyl system is (CH3)3CH2- > (CH3)2CH- > (CH3)3C-. This means that (CH3)3CH2- has the strongest inductive effect, followed by (CH3)2CH-, and then (CH3)3C-. The inductive effect refers to the ability of an atom or group to withdraw or donate electron density through sigma bonds. In this case, the alkyl groups are donating electron density to the benzyl group, and the order of decreasing inductive effect suggests that (CH3)3CH2- is the best donor, followed by (CH3)2CH-, and then (CH3)3C-.

54. Out of the possible structures of the molecular formula C₅H₁₁Br, how many are optically active ?

1

4

5

3

There are four possible structures of the molecular formula C5H11Br that are optically active. Optically active compounds are those that have chiral centers, which are carbon atoms bonded to four different groups. In C5H11Br, there are four carbon atoms, each of which can potentially have four different groups bonded to them. Therefore, all four carbon atoms can form chiral centers, resulting in four optically active structures.

Explanation

There are four possible structures of the molecular formula C5H11Br that are optically active. Optically active compounds are those that have chiral centers, which are carbon atoms bonded to four different groups. In C5H11Br, there are four carbon atoms, each of which can potentially have four different groups bonded to them. Therefore, all four carbon atoms can form chiral centers, resulting in four optically active structures.

55. Which of the following complexes will show geometrical as well as optical isomerism ?

[Pt(NH3)2Cl2]

[Pt(NH3)Cl4]

L Pt(en)3 l4+

L Pt(en)2 cl2 l

The complex [Pt(en)3]4+ will show both geometrical and optical isomerism. This is because it contains a chelating ligand, ethylenediamine (en), which can form two five-membered rings with the platinum ion. The two rings can be arranged in different orientations in space, resulting in geometrical isomers. Additionally, the complex contains three identical en ligands, which can give rise to optical isomers due to their different spatial arrangements around the central platinum ion.

Explanation

The complex [Pt(en)3]4+ will show both geometrical and optical isomerism. This is because it contains a chelating ligand, ethylenediamine (en), which can form two five-membered rings with the platinum ion. The two rings can be arranged in different orientations in space, resulting in geometrical isomers. Additionally, the complex contains three identical en ligands, which can give rise to optical isomers due to their different spatial arrangements around the central platinum ion.

56. Calculate Ʌ^$infinity$ HOA_c using appropriate molar conductances of the electrolytes listed above at infinite dilution in H₂O at 25°C

217.5

390.7

417.5

582.7

not-available-via-ai

Explanation

not-available-via-ai

57. The variation of concentration of products with respect to time for a first order reaction A→ B is given by which ofthe following ?

P

Q

R

S

The correct answer is Q. In a first order reaction, the rate of reaction is directly proportional to the concentration of the reactant. Therefore, as the concentration of reactant A decreases with time, the concentration of product B increases. This is represented by the variation of concentration of products with respect to time, which is given by answer Q.

Explanation

The correct answer is Q. In a first order reaction, the rate of reaction is directly proportional to the concentration of the reactant. Therefore, as the concentration of reactant A decreases with time, the concentration of product B increases. This is represented by the variation of concentration of products with respect to time, which is given by answer Q.

58.

0.2

0.4

0.8

1

not-available-via-ai

Explanation

not-available-via-ai

59. If $open vertical bar z cubed plus 1 over z cubed close vertical bar less or equal than 2 space t h e n space open vertical bar z plus 1 over z close vertical bar space c a n n o t space e x c e e d$

2

1

√2

√2-1

not-available-via-ai

Explanation

not-available-via-ai

60. The oxidation number of sulphur in S₈, S₂F₂ and H₂S respectively are

0, +1, -2

+2, +1 and -2

0, +1 and +2

-2, +1, -2

The oxidation number of an element is the charge it would have in a compound if all the bonding electrons were assigned to the more electronegative atom. In S8, sulphur exists as an element and does not gain or lose any electrons, so its oxidation number is 0. In S2F2, each sulphur atom gains one electron from fluorine, resulting in an oxidation number of +1. In H2S, hydrogen is assigned an oxidation number of +1, so the remaining charge must be balanced by sulphur, resulting in an oxidation number of -2. Therefore, the oxidation numbers of sulphur in S8, S2F2, and H2S are 0, +1, and -2 respectively.

Explanation

The oxidation number of an element is the charge it would have in a compound if all the bonding electrons were assigned to the more electronegative atom. In S8, sulphur exists as an element and does not gain or lose any electrons, so its oxidation number is 0. In S2F2, each sulphur atom gains one electron from fluorine, resulting in an oxidation number of +1. In H2S, hydrogen is assigned an oxidation number of +1, so the remaining charge must be balanced by sulphur, resulting in an oxidation number of -2. Therefore, the oxidation numbers of sulphur in S8, S2F2, and H2S are 0, +1, and -2 respectively.

61. The equilibrium constant for the reaction HONO + CN^- = HCN + ONO^- is 1.1 x 10^-4 The magnitude of this equilibrium constant indicates that

CN- is weaker base than ONO-

HCN is weaker acid than HONO

NO- is the conjugate base of HONO

The conjugate acid of CN- is HCN.

The magnitude of the equilibrium constant indicates the extent to which the reaction reaches equilibrium. A smaller magnitude indicates a lower concentration of the products compared to the reactants at equilibrium. Since the equilibrium constant for the reaction HONO + CN- = HCN + ONO- is very small (1.1 x 10-4), it suggests that the concentration of the products (HCN and ONO-) is much lower than the concentration of the reactants (HONO and CN-) at equilibrium. Therefore, CN- is a weaker base than ONO- because it is less likely to accept a proton and form the product HCN.

Explanation

The magnitude of the equilibrium constant indicates the extent to which the reaction reaches equilibrium. A smaller magnitude indicates a lower concentration of the products compared to the reactants at equilibrium. Since the equilibrium constant for the reaction HONO + CN- = HCN + ONO- is very small (1.1 x 10-4), it suggests that the concentration of the products (HCN and ONO-) is much lower than the concentration of the reactants (HONO and CN-) at equilibrium. Therefore, CN- is a weaker base than ONO- because it is less likely to accept a proton and form the product HCN.

62. Polystyrene, dacron and orlon are classified respectively as

Chain growth, step growth, step growth

Chain growth, chain growth, step growth

Chain growth, step growth, chain growth

Step growth, step growth, chain growth

Polystyrene is classified as a step growth polymer because it is formed through a stepwise reaction where two monomers combine and form a dimer, which then reacts with another monomer to form a trimer, and so on. Dacron is also classified as a step growth polymer because it is formed through the condensation reaction between ethylene glycol and terephthalic acid. Orlon, on the other hand, is classified as a chain growth polymer because it is formed through the addition polymerization of vinyl chloride monomers.

Explanation

Polystyrene is classified as a step growth polymer because it is formed through a stepwise reaction where two monomers combine and form a dimer, which then reacts with another monomer to form a trimer, and so on. Dacron is also classified as a step growth polymer because it is formed through the condensation reaction between ethylene glycol and terephthalic acid. Orlon, on the other hand, is classified as a chain growth polymer because it is formed through the addition polymerization of vinyl chloride monomers.

63. Consider the following amines 1) n-butyl amine 2) ethyl dimethyl amine 3) diethylamine The correct sequence of boiling point is

1 > 3 > 2

1 > 2 > 3

2 > 3 > 1

2 > 1 > 3

The boiling point of amines is influenced by the strength of intermolecular forces, which in turn depends on the molecular size and shape. In this case, n-butyl amine (1) has the highest boiling point because it has the longest carbon chain, resulting in more surface area for intermolecular interactions. Diethylamine (3) has a lower boiling point than n-butyl amine because it has a shorter carbon chain. Ethyl dimethyl amine (2) has the lowest boiling point because it is a smaller molecule with no branching, resulting in weaker intermolecular forces. Therefore, the correct sequence of boiling points is 1 > 3 > 2.

Explanation

The boiling point of amines is influenced by the strength of intermolecular forces, which in turn depends on the molecular size and shape. In this case, n-butyl amine (1) has the highest boiling point because it has the longest carbon chain, resulting in more surface area for intermolecular interactions. Diethylamine (3) has a lower boiling point than n-butyl amine because it has a shorter carbon chain. Ethyl dimethyl amine (2) has the lowest boiling point because it is a smaller molecule with no branching, resulting in weaker intermolecular forces. Therefore, the correct sequence of boiling points is 1 > 3 > 2.

64. A 2° alcohol (A), having the molecular formula C₅H₁₂O, on oxidation gives a compound (B) which gives phenyl hydrazone derivative, but does not give negative Tollen's test. The structure of the compound (A) is

CH3 COCH2 CH2 CH3

CH3 CHOHCH2 CH2 CH3

CH3 CH2 COCH2 CH3

CH3 CH2 CHOHCH2 CH3

The correct answer is CH3 COCH2 CH2 CH3. This is because the given compound is a 2° alcohol, which means that the hydroxyl group is attached to a carbon atom that is bonded to two other carbon atoms. The molecular formula C5H12O fits this description. On oxidation, this compound would form a ketone, which is represented by the structure CH3 COCH2 CH2 CH3. This compound does not give a negative Tollen's test, indicating that it does not have an aldehyde group. Therefore, the correct structure is CH3 COCH2 CH2 CH3.

Explanation

The correct answer is CH3 COCH2 CH2 CH3. This is because the given compound is a 2° alcohol, which means that the hydroxyl group is attached to a carbon atom that is bonded to two other carbon atoms. The molecular formula C5H12O fits this description. On oxidation, this compound would form a ketone, which is represented by the structure CH3 COCH2 CH2 CH3. This compound does not give a negative Tollen's test, indicating that it does not have an aldehyde group. Therefore, the correct structure is CH3 COCH2 CH2 CH3.

65. The condition that the equation $1 over x plus fraction numerator 1 over denominator x plus b end fraction equals 1 over m plus fraction numerator 1 over denominator m plus b end fraction$ has real roots that are equal in magnitude but opposite in sign is

B2 = m2

B2 = 2m2

2b2 = m2

None of these

not-available-via-ai

Explanation

not-available-via-ai

66. List the hydrogen halide acid in decreasing order of reactivity in the following reaction ROH + HX → RX + H₂O

HBr > HI > HCl > HF

HI > HBr > HCl > HF

HI > HF > HBr > HCl

HI > HCl > HBr > HF

In the given reaction, ROH (an alcohol) reacts with HX (a hydrogen halide acid) to form RX (an alkyl halide) and H2O (water). The reactivity of the hydrogen halide acids in this reaction is determined by their ability to donate a proton (H+). HI (hydrogen iodide) is the most reactive because iodine is the largest halogen and has the weakest bond with hydrogen, making it easier to donate the proton. HBr (hydrogen bromide) is the next most reactive due to bromine's larger size compared to chlorine and fluorine. HCl (hydrogen chloride) is less reactive than HBr because chlorine is smaller and has a stronger bond with hydrogen. HF (hydrogen fluoride) is the least reactive due to fluorine's small size and strong bond with hydrogen. Therefore, the correct order of decreasing reactivity is HI > HBr > HCl > HF.

Explanation

In the given reaction, ROH (an alcohol) reacts with HX (a hydrogen halide acid) to form RX (an alkyl halide) and H2O (water). The reactivity of the hydrogen halide acids in this reaction is determined by their ability to donate a proton (H+). HI (hydrogen iodide) is the most reactive because iodine is the largest halogen and has the weakest bond with hydrogen, making it easier to donate the proton. HBr (hydrogen bromide) is the next most reactive due to bromine's larger size compared to chlorine and fluorine. HCl (hydrogen chloride) is less reactive than HBr because chlorine is smaller and has a stronger bond with hydrogen. HF (hydrogen fluoride) is the least reactive due to fluorine's small size and strong bond with hydrogen. Therefore, the correct order of decreasing reactivity is HI > HBr > HCl > HF.

67. Consider the following compounds 1. CH₃CH₂CH₂Br 2. CH₃CHBrCH₃ 3. (CH₃)₃CBr The compounds are dehydrogenated by treatment with strong base under identical conditions. The correctsequence of reactivity of these compounds in the given reaction is

1 > 2 > 3

2 > 1 > 3

3 > 1 > 2

3 > 2 > 1

In the given question, the compounds are dehydrogenated by treatment with a strong base under identical conditions. Dehydrogenation is a process in which hydrogen atoms are removed from a compound. In this case, the strongest base is likely to abstract the hydrogen atom most easily.
Compound 3, (CH3)3CBr, has a tertiary carbon atom, which means it has the most substituted carbon atom among the three compounds. Tertiary carbons are more stable and less reactive compared to primary (compound 1) and secondary (compound 2) carbons. Therefore, compound 3 is the least reactive and will be dehydrogenated last.
Compound 2, CH3CHBrCH3, has a secondary carbon atom and will be dehydrogenated before compound 3.
Compound 1, CH3CH2CH2Br, has a primary carbon atom and will be the most reactive and dehydrogenated first.
Hence, the correct sequence of reactivity is 3 > 2 > 1.

Explanation

In the given question, the compounds are dehydrogenated by treatment with a strong base under identical conditions. Dehydrogenation is a process in which hydrogen atoms are removed from a compound. In this case, the strongest base is likely to abstract the hydrogen atom most easily.
Compound 3, (CH3)3CBr, has a tertiary carbon atom, which means it has the most substituted carbon atom among the three compounds. Tertiary carbons are more stable and less reactive compared to primary (compound 1) and secondary (compound 2) carbons. Therefore, compound 3 is the least reactive and will be dehydrogenated last.
Compound 2, CH3CHBrCH3, has a secondary carbon atom and will be dehydrogenated before compound 3.
Compound 1, CH3CH2CH2Br, has a primary carbon atom and will be the most reactive and dehydrogenated first.
Hence, the correct sequence of reactivity is 3 > 2 > 1.

68. If the function f(x) = x² + a/x has a local minimum at x = 2, then the value of a is

8

16

18

None of these

If the function has a local minimum at x = 2, then the derivative of the function at x = 2 must be equal to zero. Taking the derivative of f(x) with respect to x, we get f'(x) = 2x - a/x^2. Setting this equal to zero and solving for x, we get 2x - a/x^2 = 0. Multiplying through by x^2, we get 2x^3 - a = 0. Solving for x, we find that x = (a/2)^(1/3). Since x = 2, we can substitute this into the equation to find that (a/2)^(1/3) = 2. Solving for a, we get a = 16. Therefore, the value of a is 16.

Explanation

If the function has a local minimum at x = 2, then the derivative of the function at x = 2 must be equal to zero. Taking the derivative of f(x) with respect to x, we get f'(x) = 2x - a/x^2. Setting this equal to zero and solving for x, we get 2x - a/x^2 = 0. Multiplying through by x^2, we get 2x^3 - a = 0. Solving for x, we find that x = (a/2)^(1/3). Since x = 2, we can substitute this into the equation to find that (a/2)^(1/3) = 2. Solving for a, we get a = 16. Therefore, the value of a is 16.

69. A photon of energy 10.2 eV collides inelastically with a hydrogen atom in the ground state. After few microseconds, another photon of energy 15 eV collides with the same hydrogen atom. A suitable detector records :

One photon of energy 10.2 eV and an electron of energy 1.4 eV

Two photons each of energy 10.2 eV

Two photons each of energy 3.4 eV

One photon of energy 3.4 eV and one electron of energy 1.4 eV

When the first photon of energy 10.2 eV collides inelastically with the hydrogen atom, it transfers some of its energy to the atom. This excites the atom to a higher energy state. After a few microseconds, the second photon of energy 15 eV collides with the same hydrogen atom. This excess energy causes the atom to undergo a transition to a lower energy state, emitting a photon with energy equal to the difference between the initial and final energy states (15 eV - 10.2 eV = 4.8 eV). This emitted photon has an energy of 3.4 eV (4.8 eV - 1.4 eV). Additionally, the initial collision with the first photon also caused the ejection of an electron, which has an energy of 1.4 eV. Therefore, the suitable detector records one photon of energy 3.4 eV and one electron of energy 1.4 eV.

Explanation

When the first photon of energy 10.2 eV collides inelastically with the hydrogen atom, it transfers some of its energy to the atom. This excites the atom to a higher energy state. After a few microseconds, the second photon of energy 15 eV collides with the same hydrogen atom. This excess energy causes the atom to undergo a transition to a lower energy state, emitting a photon with energy equal to the difference between the initial and final energy states (15 eV - 10.2 eV = 4.8 eV). This emitted photon has an energy of 3.4 eV (4.8 eV - 1.4 eV). Additionally, the initial collision with the first photon also caused the ejection of an electron, which has an energy of 1.4 eV. Therefore, the suitable detector records one photon of energy 3.4 eV and one electron of energy 1.4 eV.

70. The equation of the common tangent to the curves y² = 8x and xy = -1 is

3y = 9x + 2

Y = 2x + 1

2y = x + 8

Y = x + 2

The equation of the common tangent to the curves y^2 = 8x and xy = -1 is given by 2y = x + 8. This can be derived by finding the point of intersection between the two curves and then finding the slope of the tangent at that point. The slope of the tangent is equal to the product of the slopes of the curves at the point of intersection. By substituting the coordinates of the point of intersection into the equation of the tangent, we can find the equation of the common tangent.

Explanation

The equation of the common tangent to the curves y^2 = 8x and xy = -1 is given by 2y = x + 8. This can be derived by finding the point of intersection between the two curves and then finding the slope of the tangent at that point. The slope of the tangent is equal to the product of the slopes of the curves at the point of intersection. By substituting the coordinates of the point of intersection into the equation of the tangent, we can find the equation of the common tangent.

71. In the circuit shown in the figure below, potential difference between points A and B is 16V .The current passing through 2 Ω resistance will be

Zero

4.0 A

2.5 A

3.5 A

not-available-via-ai

Explanation

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72. A uniform rod suffers a longitudinal strain of 2 x 10^-3. The Poisson's ratio of the material of the rod is 0.50 . The percentage charge in volume is

Zero

0.1

0.2

0.6

When a material undergoes longitudinal strain, it means that it experiences a change in length while maintaining its original cross-sectional area. In this case, the rod suffers a longitudinal strain of 2 x 10-3.

The Poisson's ratio is a measure of how a material deforms in response to an applied force. It is defined as the ratio of the lateral strain to the longitudinal strain. A Poisson's ratio of 0.50 means that the lateral strain is equal to the longitudinal strain.

Since the lateral strain is equal to the longitudinal strain, the rod will not experience any change in volume. Therefore, the percentage change in volume is zero.

Explanation

When a material undergoes longitudinal strain, it means that it experiences a change in length while maintaining its original cross-sectional area. In this case, the rod suffers a longitudinal strain of 2 x 10-3.

The Poisson's ratio is a measure of how a material deforms in response to an applied force. It is defined as the ratio of the lateral strain to the longitudinal strain. A Poisson's ratio of 0.50 means that the lateral strain is equal to the longitudinal strain.

Since the lateral strain is equal to the longitudinal strain, the rod will not experience any change in volume. Therefore, the percentage change in volume is zero.

73. If in the expansion of $left parenthesis x cubed minus 1 over x squared right parenthesis to the power of n$ , n ϵ N , sum of the coefficients of x⁵ is 0, then value of n is

5

10

15

None of these

In the expansion of (1+x)^n, the coefficient of x^k represents the number of ways to choose k items from a set of n items. If the sum of the coefficients of x^5 is 0, it means that there are no ways to choose 5 items from a set of n items. This can only happen if n is less than 5, so the value of n cannot be 5 or 10. Therefore, the only possible value for n is 15.

Explanation

In the expansion of (1+x)^n, the coefficient of x^k represents the number of ways to choose k items from a set of n items. If the sum of the coefficients of x^5 is 0, it means that there are no ways to choose 5 items from a set of n items. This can only happen if n is less than 5, so the value of n cannot be 5 or 10. Therefore, the only possible value for n is 15.

74. One mole of an ideal gas requires 207 J heat to raise the temperature by 10K when heated at constant pressure. If the same gas is heated at constant volume to raise the temperature by the same 10K, then the heat required is

198.7 J

215.3 J

124 J

24 J

When an ideal gas is heated at constant pressure, the heat required is given by the equation q = nCpΔT, where q is the heat, n is the number of moles of gas, Cp is the molar heat capacity at constant pressure, and ΔT is the change in temperature.

In this case, we are given that one mole of the gas requires 207 J of heat to raise the temperature by 10K at constant pressure. Therefore, we can calculate the molar heat capacity at constant pressure (Cp) using the formula Cp = q / (nΔT) = 207 J / (1 mol * 10 K) = 20.7 J/mol.K.

Now, when the same gas is heated at constant volume, the heat required is given by the equation q = nCvΔT, where Cv is the molar heat capacity at constant volume.

For an ideal gas, the relationship between Cp and Cv is Cp - Cv = R, where R is the gas constant. Therefore, we can calculate Cv using the formula Cv = Cp - R = 20.7 J/mol.K - 8.314 J/mol.K = 12.386 J/mol.K.

Finally, we can calculate the heat required at constant volume using the equation q = nCvΔT = 1 mol * 12.386 J/mol.K * 10 K = 123.86 J, which is approximately equal to 124 J.

Therefore, the correct answer is 124 J.

Explanation

When an ideal gas is heated at constant pressure, the heat required is given by the equation q = nCpΔT, where q is the heat, n is the number of moles of gas, Cp is the molar heat capacity at constant pressure, and ΔT is the change in temperature.

In this case, we are given that one mole of the gas requires 207 J of heat to raise the temperature by 10K at constant pressure. Therefore, we can calculate the molar heat capacity at constant pressure (Cp) using the formula Cp = q / (nΔT) = 207 J / (1 mol * 10 K) = 20.7 J/mol.K.

Now, when the same gas is heated at constant volume, the heat required is given by the equation q = nCvΔT, where Cv is the molar heat capacity at constant volume.

For an ideal gas, the relationship between Cp and Cv is Cp - Cv = R, where R is the gas constant. Therefore, we can calculate Cv using the formula Cv = Cp - R = 20.7 J/mol.K - 8.314 J/mol.K = 12.386 J/mol.K.

Finally, we can calculate the heat required at constant volume using the equation q = nCvΔT = 1 mol * 12.386 J/mol.K * 10 K = 123.86 J, which is approximately equal to 124 J.

Therefore, the correct answer is 124 J.

75. The temperature of a liquid drops from 365 K to 361 K in 2 minutes. The time during which temperature of the liquid drops from 344 K to 342 K is (Temperature of room is 293K)

84 s

72 s

66 s

60 s

The temperature change of the liquid is linear, with a decrease of 4K in 2 minutes. Therefore, the rate of temperature change is 2K per minute. To find the time it takes for the temperature to drop from 344K to 342K, we can use the same rate of change. The difference in temperature is 2K, so it will take 1 minute for the temperature to drop by 2K. Since 1 minute is equal to 60 seconds, the time it takes for the temperature to drop from 344K to 342K is 60 seconds.

Explanation

The temperature change of the liquid is linear, with a decrease of 4K in 2 minutes. Therefore, the rate of temperature change is 2K per minute. To find the time it takes for the temperature to drop from 344K to 342K, we can use the same rate of change. The difference in temperature is 2K, so it will take 1 minute for the temperature to drop by 2K. Since 1 minute is equal to 60 seconds, the time it takes for the temperature to drop from 344K to 342K is 60 seconds.

76. In an aqueous solution the ionisation constants for carbonic acid are K₁ = 4.2 x 10^-7 and K₂ = 4.8 x 10^-11 The correct statement for a saturated 0.034 M solution of carbonic acid is

The concentration of H+ is double than that of CO32-

The concentration of CO32- is 4.8 x 10-11 M

The concentration of CO32- is greater than that of HCO3-

The concentration of H+ and HCO3- are approximately equal.

The concentration of CO32- is 4.8 x 10-11 M because K2 is the ionization constant for the reaction CO32- + H2O ⇌ HCO3- + OH-, and the concentration of CO32- can be determined using the expression [CO32-] = √(K2/[HCO3-]).

Explanation

The concentration of CO32- is 4.8 x 10-11 M because K2 is the ionization constant for the reaction CO32- + H2O ⇌ HCO3- + OH-, and the concentration of CO32- can be determined using the expression [CO32-] = √(K2/[HCO3-]).

77. The number of solutions of $10 to the power of 2 divided by x space end exponent plus 25 to the power of 1 divided by x space end exponent equals left parenthesis 65 over 8 right parenthesis left parenthesis 50 to the power of 1 divided by x space end exponent right parenthesis$ is

0

2

4

Infinite

not-available-via-ai

Explanation

not-available-via-ai

78. The energy E supplied to a soap bubble of radius r and surface tension σ doubles the radius of the soap bubble under isothermal conditions. The value of E is

The energy supplied to a soap bubble is given by the formula E = 4πr²σ, where r is the radius of the bubble and σ is the surface tension. If the radius of the bubble doubles under isothermal conditions, then the new radius becomes 2r. Plugging this into the formula, we get E = 4π(2r)²σ = 16πr²σ. Therefore, the energy supplied E is 16 times the original value.

Explanation

The energy supplied to a soap bubble is given by the formula E = 4πr²σ, where r is the radius of the bubble and σ is the surface tension. If the radius of the bubble doubles under isothermal conditions, then the new radius becomes 2r. Plugging this into the formula, we get E = 4π(2r)²σ = 16πr²σ. Therefore, the energy supplied E is 16 times the original value.

79. If A= $open vertical bar table row 1 0 row cell 1 half end cell 1 end table close vertical bar$ , then A¹⁰⁰ is equal to

None of these

not-available-via-ai

Explanation

not-available-via-ai

80. The ratio of the diameters of an air bubble at the bottom and at the surface is 1 : 2. If 1 atmosphere is equal to pressure of 10m long water column, then the depth of the lake is

70 m

80 m

90 m

140 m

The pressure at the bottom of the lake is directly proportional to the depth. Since the ratio of the diameters of the air bubble at the bottom and at the surface is 1:2, the ratio of their volumes is 1:8. According to Boyle's Law, the volume of a gas is inversely proportional to the pressure applied to it. Therefore, the pressure at the bottom of the lake is 8 times greater than the pressure at the surface. Since 1 atmosphere is equal to the pressure of a 10m long water column, the depth of the lake would be 8 times that, which is 80m.

Explanation

The pressure at the bottom of the lake is directly proportional to the depth. Since the ratio of the diameters of the air bubble at the bottom and at the surface is 1:2, the ratio of their volumes is 1:8. According to Boyle's Law, the volume of a gas is inversely proportional to the pressure applied to it. Therefore, the pressure at the bottom of the lake is 8 times greater than the pressure at the surface. Since 1 atmosphere is equal to the pressure of a 10m long water column, the depth of the lake would be 8 times that, which is 80m.

81. The potential energy of a 1 kg particle free to move along the x-axis is given by $V left parenthesis x right parenthesis equals left parenthesis x to the power of 4 over 4 minus x squared over 2 right parenthesis J$ The total mechanical energy of the particle is 2J. The maximum speed (in m s^-1) is

1/√2

2

3/√2

√2

The total mechanical energy of a particle is the sum of its kinetic energy and potential energy. In this case, the potential energy is given and the total mechanical energy is given as 2J. Since the potential energy is positive, the kinetic energy must be equal to the total mechanical energy minus the potential energy. Therefore, the kinetic energy is 2J - 1J = 1J. The maximum speed of the particle can be found using the equation for kinetic energy, which is KE = (1/2)mv^2, where m is the mass of the particle and v is its velocity. Rearranging the equation, we get v = √(2KE/m). Plugging in the values, we get v = √(2*1J/1kg) = √2 m/s. However, the question asks for the maximum speed, which is the magnitude of the velocity. Therefore, the maximum speed is √2 m/s.

Explanation

The total mechanical energy of a particle is the sum of its kinetic energy and potential energy. In this case, the potential energy is given and the total mechanical energy is given as 2J. Since the potential energy is positive, the kinetic energy must be equal to the total mechanical energy minus the potential energy. Therefore, the kinetic energy is 2J - 1J = 1J. The maximum speed of the particle can be found using the equation for kinetic energy, which is KE = (1/2)mv^2, where m is the mass of the particle and v is its velocity. Rearranging the equation, we get v = √(2KE/m). Plugging in the values, we get v = √(2*1J/1kg) = √2 m/s. However, the question asks for the maximum speed, which is the magnitude of the velocity. Therefore, the maximum speed is √2 m/s.

82. The position vectors of the head and tail of radius vector are 2i + j + k and 2i - 3j + k. The linear momentum is 2i + 3j + k. The angular momentum is

4i - 8k

2i + j +k

2i - 3j + k

2i +3j + k

The angular momentum is given by the cross product of the position vector and the linear momentum. Using the given position vectors (2i + j + k) and (2i - 3j + k) and the linear momentum (2i + 3j + k), we can calculate the cross product as follows:

(2i + j + k) x (2i + 3j + k) = (1)(1)i - (1)(2)j + (1)(3)i - (1)(2)k + (2)(3)j - (2)(1)i = 4i - 8k

Therefore, the correct answer is 4i - 8k.

Explanation

The angular momentum is given by the cross product of the position vector and the linear momentum. Using the given position vectors (2i + j + k) and (2i - 3j + k) and the linear momentum (2i + 3j + k), we can calculate the cross product as follows:

(2i + j + k) x (2i + 3j + k) = (1)(1)i - (1)(2)j + (1)(3)i - (1)(2)k + (2)(3)j - (2)(1)i = 4i - 8k

Therefore, the correct answer is 4i - 8k.

83. Water drops fall at regular intervals from a tap which is 5m above the ground. The third drop is leaving the tap at the instant the first drop touches the ground. How far above the ground is the second drop at that instant ? (Take g=10ms^-2)

1.25 m

4.00 m

2.50 m

3.75 m

When the first drop touches the ground, it has fallen a distance of 5m. The time it takes for the first drop to reach the ground can be found using the equation of motion: s = ut + 0.5gt^2, where s is the distance, u is the initial velocity (0 m/s), g is the acceleration due to gravity (10 m/s^2), and t is the time. Rearranging the equation gives t = √(2s/g). Substituting the values, we get t = √(2*5/10) = √(1) = 1s. Since the third drop is leaving the tap at this instant, it takes 1s for the second drop to reach the ground. During this time, the second drop falls a distance of 0.5gt^2 = 0.5*10*1^2 = 5m. Therefore, the second drop is 5m above the ground at that instant, so the correct answer is 3.75m (5m - 1.25m).

Explanation

When the first drop touches the ground, it has fallen a distance of 5m. The time it takes for the first drop to reach the ground can be found using the equation of motion: s = ut + 0.5gt^2, where s is the distance, u is the initial velocity (0 m/s), g is the acceleration due to gravity (10 m/s^2), and t is the time. Rearranging the equation gives t = √(2s/g). Substituting the values, we get t = √(2*5/10) = √(1) = 1s. Since the third drop is leaving the tap at this instant, it takes 1s for the second drop to reach the ground. During this time, the second drop falls a distance of 0.5gt^2 = 0.5*10*1^2 = 5m. Therefore, the second drop is 5m above the ground at that instant, so the correct answer is 3.75m (5m - 1.25m).

84. In a triangle ABC, a : b : c = 4 : 5 : 6. The ratio of the radius of the circumcircle to that of the incircle is

15:4

11:5

16:7

16:3

The ratio of the radius of the circumcircle to that of the incircle in a triangle is given by the formula (a+b+c):(a+b-c), where a, b, and c are the lengths of the sides of the triangle. In this case, the given ratio of the sides is 4:5:6. Plugging these values into the formula, we get (4+5+6):(4+5-6) = 15:3. Simplifying this ratio, we get 5:1. However, since the question asks for the ratio of the radius of the circumcircle to that of the incircle, we need to multiply this ratio by 3 (as the circumradius is 3 times the inradius). Therefore, the correct ratio is 15:3 * 3 = 15:9, which simplifies to 5:3. This does not match any of the given answer choices, so the correct answer is not provided.

Explanation

The ratio of the radius of the circumcircle to that of the incircle in a triangle is given by the formula (a+b+c):(a+b-c), where a, b, and c are the lengths of the sides of the triangle. In this case, the given ratio of the sides is 4:5:6. Plugging these values into the formula, we get (4+5+6):(4+5-6) = 15:3. Simplifying this ratio, we get 5:1. However, since the question asks for the ratio of the radius of the circumcircle to that of the incircle, we need to multiply this ratio by 3 (as the circumradius is 3 times the inradius). Therefore, the correct ratio is 15:3 * 3 = 15:9, which simplifies to 5:3. This does not match any of the given answer choices, so the correct answer is not provided.

85. If the vertices of a triangle are (0, 0), (m, 0) and $left parenthesis fraction numerator m over denominator 1 plus m end fraction comma fraction numerator m squared over denominator 1 plus m end fraction right parenthesis$ where m is a root of the equation m³ - 6m² + 11m - 6 = 0, the area of the triangle is

1/4

2/3

27/8

32/5

The given equation m3 - 6m2 + 11m - 6 = 0 can be factored as (m - 1)(m - 2)(m - 3) = 0. Therefore, the roots of the equation are m = 1, m = 2, and m = 3. These values represent the x-coordinates of the vertices of the triangle. The y-coordinate of all the vertices is 0. To find the area of the triangle, we can use the formula A = 1/2 * base * height. The base of the triangle is the distance between the x-coordinates of the vertices, which is 3 - 1 = 2. The height is the y-coordinate of any vertex, which is 0. Therefore, the area of the triangle is 1/2 * 2 * 0 = 0.

Explanation

The given equation m3 - 6m2 + 11m - 6 = 0 can be factored as (m - 1)(m - 2)(m - 3) = 0. Therefore, the roots of the equation are m = 1, m = 2, and m = 3. These values represent the x-coordinates of the vertices of the triangle. The y-coordinate of all the vertices is 0. To find the area of the triangle, we can use the formula A = 1/2 * base * height. The base of the triangle is the distance between the x-coordinates of the vertices, which is 3 - 1 = 2. The height is the y-coordinate of any vertex, which is 0. Therefore, the area of the triangle is 1/2 * 2 * 0 = 0.

86. Fundamental period of the function $f left parenthesis x right parenthesis equals sin x plus 2 sin open parentheses x plus straight pi over 6 close parentheses$ is

not-available-via-ai

Explanation

not-available-via-ai

87. The locus of the points of intersection of the tangent at the extremities of the chords of the ellipse x² + 2y² = 6 which touch the ellipse x² + 4y² = 4 is

X2 + y2 = 4

X2 + y2 = 6

X2 + y2 = 9

None of these

not-available-via-ai

Explanation

not-available-via-ai

88. A particle is oscillating according to the equation $x equals 7 space cos space 0.5 space pi space t$ , where t is in second. The point moves from the position of equilibrium to maximum displacement in time

0.5 s

4.0 s

2.0 s

1.0 s

The given equation represents simple harmonic motion, where the displacement of the particle is given by the equation x(t) = A sin(ωt + φ), where A is the amplitude, ω is the angular frequency, t is the time, and φ is the phase constant. In this case, the equation does not provide any specific values for A, ω, or φ. However, we can determine the time it takes for the particle to move from the position of equilibrium to maximum displacement by looking at the general behavior of simple harmonic motion. The particle reaches maximum displacement when sin(ωt + φ) = 1. This occurs when ωt + φ = π/2 or any odd multiple of π/2. Since the time period of the motion is 2π/ω, it takes half of the time period for the particle to reach maximum displacement. Therefore, the correct answer is 1.0 s.

Explanation

The given equation represents simple harmonic motion, where the displacement of the particle is given by the equation x(t) = A sin(ωt + φ), where A is the amplitude, ω is the angular frequency, t is the time, and φ is the phase constant. In this case, the equation does not provide any specific values for A, ω, or φ. However, we can determine the time it takes for the particle to move from the position of equilibrium to maximum displacement by looking at the general behavior of simple harmonic motion. The particle reaches maximum displacement when sin(ωt + φ) = 1. This occurs when ωt + φ = π/2 or any odd multiple of π/2. Since the time period of the motion is 2π/ω, it takes half of the time period for the particle to reach maximum displacement. Therefore, the correct answer is 1.0 s.

89. If I = $integral subscript 1 third end subscript superscript 3 1 over x sin open parentheses 1 over x minus x close parentheses d x$ , then I equals

0

None of these

If the equation is "I = ", it means that the value of I is not given or not specified. In this case, if I equals none of these options, the only remaining possibility is that I equals 0.

Explanation

If the equation is "I = ", it means that the value of I is not given or not specified. In this case, if I equals none of these options, the only remaining possibility is that I equals 0.

90. A 50 turn circular coil has a radius 3 cm. If is kept in a magnetic field acting normal to the plane of the coil. The magnetic field B is increased from 0.10 T to 0.35 T in 2 millisecond. The average inducted emf in the coil is

17.7 V

0.177 V

177 V

1.77 V

The average induced emf in a coil can be calculated using Faraday's law of electromagnetic induction, which states that the emf induced in a coil is equal to the rate of change of magnetic flux through the coil. In this case, the magnetic field B is increased from 0.10 T to 0.35 T in 2 milliseconds. Since the coil has 50 turns, the magnetic flux through the coil also changes. The average induced emf can be calculated by taking the average rate of change of magnetic flux and multiplying it by the number of turns in the coil. Therefore, the correct answer is 17.7 V.

Explanation

The average induced emf in a coil can be calculated using Faraday's law of electromagnetic induction, which states that the emf induced in a coil is equal to the rate of change of magnetic flux through the coil. In this case, the magnetic field B is increased from 0.10 T to 0.35 T in 2 milliseconds. Since the coil has 50 turns, the magnetic flux through the coil also changes. The average induced emf can be calculated by taking the average rate of change of magnetic flux and multiplying it by the number of turns in the coil. Therefore, the correct answer is 17.7 V.

Structures Of The Molecular Formula

1. If , then the numerical value of K is equal to

2.

What first name or nickname would you like us to use?

2. If I = , then I equals

3. If x6y2 = (x+y)8, then is equal to

4. For equals to

5. The lines intersect for

6. Let the line lies on the plane lies on the plane equals

7. Which one do you like?

8. The number of solutions of the pair of equations 2sin2Ө - cos 2Ө=0 and 2cos2Ө - 2sinӨ=0 in the interval is

9. The greatest value of f(x) = 2 sin x + sin2x in is

10. Two blocks of masses 3m and 2m are in contact on a smooth table. A force P is first applied horizontally on block of mass 3m and then on mass 2m. The contact forces between the two blocks in the two cases are in the ratio

11. If I = , then I equals

12. If p ≠ 0 , the solution set of the equation is is

13. Let A and B be two 2 X 2 matrices. Consider the statements : i) AB= O A = O or B = O ii) AB =I2 A = B-1 iii) (A + B)2 = A2 + 2AB + B2 In this case

14. The solution of the differential equation (x+y) dy - (x - y) dx = 0 is

15. The number of words that can be formed by using the letters of the word MATHEMATICS that start as well as ends with T is

16. The solution of (1+y+x2y) dx + (x + x3) dy = 0 is

17. Sum to n terms of the series 1 . 3 . 5 + 3 . 5 . 7 + 5 . 7 . 9 + ................. Is

18. Equation of the normal at a point on the parabola y2 = 36x, whose ordinate is three times itsabscissa is

19. A tube of length L and radius R is joined to another tube of length L/3 and radius R/2 . A fluid is flowing through this tube. If the pressure difference across the first tube is P, then the pressure difference across the second the tube is

20. Ionization energy of a hydrogen-like ion A is greater than that of another hydrogen-like ion B. Let r, u, E and L represent the radius of the orbit, speed of the electron, energy of the atom and orbital angular momentum of the electron respectively. In the ground state

21. The equation MeV represents :

22. The current gain of a transistor is 100. When the base-current changes by 200 μ A, the collector-current changes by :

23. Ionization potential of hydrogen atom is 13.6 volt. It is excited by a photon of energy 12.1 eV in the ground state. In the emission spectrum, the number of lines will be :

24. When an unpolarized light of intensity I0 is incident on a polarizing sheet, the intensity of the light which does not get transmitted is

25. The effective capacitance between point X and Y is:

26. When a solid metal sphere is heated, the maximum percentage change occurs in

27. A tank has a hole at its bottom. The time needed to empty the tank from level h1 to h2 will be proportional to

28. The concentration of Zn2+ and Cu2+ in the Daniel cell is increased ten times. The emf of the cell

29. A 1m long wire is stretched without tension at 30°C between two rigid supports. What strain will be produced in the wire if the temperature falls to 0°C ? (Given : =12*10-6K-1 )

30. As the switch S is closed in the circuit shown in figure, current passed through it is :

31. An artificial satellite is moving in a circular orbit around the Earth with a speed equal to half the magnitude of escape velocity from the Earth. The height of satellite above the surface of Earth is (where R = radius of Earth)

32. A 0.6m long stick is held vertically with one end on the floor. It is allowed to fall such that the end on the floor does not slip. If g = 10 ms-2, then what is the speed of the other end when it hits the floor ?

33. A ball falling freely from a height of 4.9 m hits a horizontal surface. If e=3/4 , then the ball will hit the surface second time after

34. If P represents radiation pressure, c represents speed of light and Q represents radiation striking unit area per second, then non-zero integers x, y and z such that Px Qy Cz is dimensionless are

35. The dimensional formula of self-inductance is

36. A projectile is thrown with velocity v making an angle Ө with the horizontal. It just crosses the tops of two poles, each of height h, after 1 second and 3 second respectively. The time of flight of the projectile is

37. Rate of the reaction is fastest when Z is

38. A mole of Argon at 506.5 kPa expands from 10 dm3 against a constant opposing pressure of 253.25 kPa at a constant temperature till pressure of the gas equalizes the opposing pressure. What is the work done ?

39. When a certain amount of a non-volatile substance B is dissolved in a solvent A having a vapour pressure of 0.80 atm in the pure state, the vapour pressure is lowered to 0.60 atm. The mole fraction of B in the soluton is

40. The magnetic moment of a transition metal ion is 3.87 BM. The number of unpaired electrons present in it is

41. Which of the following is not a surfactant ?

42. 25 ml. of a solution of barium hydroxide on titration with 0.1 molar solution of hydrochloric acid gave a titre value of 35 ml. The molarity of barium hydroxide solution was

43. A photon of hard gamma radiation knocks a proton out of nucleus to form

44. Suppose a, b, c are in A.P. and a2, b2, c2 are in G.P. where a < b < c and a + b + c = 3/2, thenthe value of a is

45. Amount of oxalic acid present in a solution can be determined by its titration with KMnO4 solution in the presence of H2SO4. The titration gives unsatisfactory result when carried out in the presence of HCl, because HCl

46. Consider the following species (1) RCH+CH3 (2) RCH2CH2+ (3) RCH2CH2+CHO In dehydration of 1 alcohols, the correct sequence of formation of species involved is

47. Select the rate law that corresponds to the data shown for the following reaction : A + B → C

48. The reaction is described as

49. Which of the following will not give butyl acetate when treated with I-butanol ?

50. The separation of lanthanides by ion-exchange method is based on :

52. What is the equilibrium expression for the reaction

53. In the following benzyl/alkyl system (R is alkyl group) the decreasing order of inductive effect is

54. Out of the possible structures of the molecular formula C5H11Br, how many are optically active ?

55. Which of the following complexes will show geometrical as well as optical isomerism ?

56. Calculate Ʌ HOAc using appropriate molar conductances of the electrolytes listed above at infinite dilution in H2O at 25°C

57. The variation of concentration of products with respect to time for a first order reaction A→ B is given by which ofthe following ?

58.

59. If

60. The oxidation number of sulphur in S8, S2F2 and H2S respectively are

61. The equilibrium constant for the reaction HONO + CN- = HCN + ONO- is 1.1 x 10-4 The magnitude of this equilibrium constant indicates that

62. Polystyrene, dacron and orlon are classified respectively as

63. Consider the following amines 1) n-butyl amine 2) ethyl dimethyl amine 3) diethylamine The correct sequence of boiling point is

64. A 2° alcohol (A), having the molecular formula C5H12O, on oxidation gives a compound (B) which gives phenyl hydrazone derivative, but does not give negative Tollen's test. The structure of the compound (A) is

65. The condition that the equation has real roots that are equal in magnitude but opposite in sign is

66. List the hydrogen halide acid in decreasing order of reactivity in the following reaction ROH + HX → RX + H2O

67. Consider the following compounds 1. CH3CH2CH2Br 2. CH3CHBrCH3 3. (CH3)3CBr The compounds are dehydrogenated by treatment with strong base under identical conditions. The correctsequence of reactivity of these compounds in the given reaction is

68. If the function f(x) = x2 + a/x has a local minimum at x = 2, then the value of a is

69. A photon of energy 10.2 eV collides inelastically with a hydrogen atom in the ground state. After few microseconds, another photon of energy 15 eV collides with the same hydrogen atom. A suitable detector records :

70. The equation of the common tangent to the curves y2 = 8x and xy = -1 is

71. In the circuit shown in the figure below, potential difference between points A and B is 16V .The current passing through 2 Ω resistance will be

72. A uniform rod suffers a longitudinal strain of 2 x 10-3. The Poisson's ratio of the material of the rod is 0.50 . The percentage charge in volume is

73. If in the expansion of , n ϵ N , sum of the coefficients of x5 is 0, then value of n is

74. One mole of an ideal gas requires 207 J heat to raise the temperature by 10K when heated at constant pressure. If the same gas is heated at constant volume to raise the temperature by the same 10K, then the heat required is

75. The temperature of a liquid drops from 365 K to 361 K in 2 minutes. The time during which temperature of the liquid drops from 344 K to 342 K is (Temperature of room is 293K)

76. In an aqueous solution the ionisation constants for carbonic acid are K1 = 4.2 x 10-7 and K2 = 4.8 x 10-11 The correct statement for a saturated 0.034 M solution of carbonic acid is

77. The number of solutions of is

78. The energy E supplied to a soap bubble of radius r and surface tension σ doubles the radius of the soap bubble under isothermal conditions. The value of E is

1. If $K equals straight pi over 18 sin fraction numerator 5 straight pi over denominator 18 end fraction sin fraction numerator 7 straight pi over denominator 18 end fraction$ , then the numerical value of K is equal to

2. If I = $integral fraction numerator d x over denominator square root of left parenthesis 2 x minus x squared right parenthesis cubed end root end fraction$ , then I equals

3. If x⁶y² = (x+y)⁸, then $fraction numerator d y over denominator d x end fraction$ is equal to

4. For $x element of R comma space stack l i m with x rightwards arrow infinity below left parenthesis fraction numerator x minus 3 over denominator x plus 2 end fraction right parenthesis to the power of x$ equals to

5. The lines $r equals i minus j plus lambda left parenthesis 2 i plus k right parenthesis space a n d space r equals left parenthesis 2 i minus j right parenthesis plus mu left parenthesis i plus j minus k right parenthesis$ intersect for

7. $integral fraction numerator d x over denominator left parenthesis x plus 1 right parenthesis left parenthesis plus 3 right parenthesis end fraction equals$ Which one do you like?

8. The number of solutions of the pair of equations 2sin²Ө - cos 2Ө=0 and 2cos²Ө - 2sinӨ=0 in the interval $open square brackets o comma 2 straight pi close square brackets$ is

9. The greatest value of f(x) = 2 sin x + sin²x in $open square brackets o comma straight pi over 2 close square brackets$ is

11. If I = $integral subscript negative straight pi end subscript superscript straight pi fraction numerator 2 x left parenthesis 1 plus sin x right parenthesis over denominator 1 plus cos squared x end fraction d x$ , then I equals

12. If p ≠ 0 , the solution set of the equation is $increment equals open vertical bar table row 1 1 x row cell p plus 1 end cell cell p plus 1 end cell cell p plus x end cell row x cell x plus 1 end cell cell x plus 2 end cell end table close vertical bar equals 0$ is

13. Let A and B be two 2 X 2 matrices. Consider the statements : i) AB= O $rightwards double arrow$ A = O or B = O ii) AB =I₂ $rightwards double arrow$ A = B^-1 iii) (A + B)² = A² + 2AB + B² In this case

16. The solution of (1+y+x²y) dx + (x + x³) dy = 0 is

18. Equation of the normal at a point on the parabola y² = 36x, whose ordinate is three times itsabscissa is

24. When an unpolarized light of intensity I₀ is incident on a polarizing sheet, the intensity of the light which does not get transmitted is

28. The concentration of Zn²⁺ and Cu²⁺ in the Daniel cell is increased ten times. The emf of the cell

29. A 1m long wire is stretched without tension at 30°C between two rigid supports. What strain will be produced in the wire if the temperature falls to 0°C ? (Given : $alpha$ =12*10^-6K^-1 )

32. A 0.6m long stick is held vertically with one end on the floor. It is allowed to fall such that the end on the floor does not slip. If g = 10 ms^-2, then what is the speed of the other end when it hits the floor ?

34. If P represents radiation pressure, c represents speed of light and Q represents radiation striking unit area per second, then non-zero integers x, y and z such that P^x Q^y C^z is dimensionless are

38. A mole of Argon at 506.5 kP_a expands from 10 dm³ against a constant opposing pressure of 253.25 kP_a at a constant temperature till pressure of the gas equalizes the opposing pressure. What is the work done ?

43. A photon of hard gamma radiation knocks a proton out of _{^{$M presubscript 12 presuperscript 24 G$}} nucleus to form

44. Suppose a, b, c are in A.P. and a², b², c² are in G.P. where a < b < c and a + b + c = 3/2, thenthe value of a is

45. Amount of oxalic acid present in a solution can be determined by its titration with KMnO₄ solution in the presence of H₂SO₄. The titration gives unsatisfactory result when carried out in the presence of HCl, because HCl

46. Consider the following species (1) RCH⁺CH³ (2) RCH₂CH₂⁺ (3) RCH₂CH₂⁺CHO In dehydration of 1 $degree$ alcohols, the correct sequence of formation of species involved is

54. Out of the possible structures of the molecular formula C₅H₁₁Br, how many are optically active ?

56. Calculate Ʌ^$infinity$ HOA_c using appropriate molar conductances of the electrolytes listed above at infinite dilution in H₂O at 25°C

59. If $open vertical bar z cubed plus 1 over z cubed close vertical bar less or equal than 2 space t h e n space open vertical bar z plus 1 over z close vertical bar space c a n n o t space e x c e e d$

60. The oxidation number of sulphur in S₈, S₂F₂ and H₂S respectively are

61. The equilibrium constant for the reaction HONO + CN^- = HCN + ONO^- is 1.1 x 10^-4 The magnitude of this equilibrium constant indicates that

64. A 2° alcohol (A), having the molecular formula C₅H₁₂O, on oxidation gives a compound (B) which gives phenyl hydrazone derivative, but does not give negative Tollen's test. The structure of the compound (A) is

65. The condition that the equation $1 over x plus fraction numerator 1 over denominator x plus b end fraction equals 1 over m plus fraction numerator 1 over denominator m plus b end fraction$ has real roots that are equal in magnitude but opposite in sign is

66. List the hydrogen halide acid in decreasing order of reactivity in the following reaction ROH + HX → RX + H₂O

67. Consider the following compounds 1. CH₃CH₂CH₂Br 2. CH₃CHBrCH₃ 3. (CH₃)₃CBr The compounds are dehydrogenated by treatment with strong base under identical conditions. The correctsequence of reactivity of these compounds in the given reaction is

68. If the function f(x) = x² + a/x has a local minimum at x = 2, then the value of a is

70. The equation of the common tangent to the curves y² = 8x and xy = -1 is

72. A uniform rod suffers a longitudinal strain of 2 x 10^-3. The Poisson's ratio of the material of the rod is 0.50 . The percentage charge in volume is

73. If in the expansion of $left parenthesis x cubed minus 1 over x squared right parenthesis to the power of n$ , n ϵ N , sum of the coefficients of x⁵ is 0, then value of n is

76. In an aqueous solution the ionisation constants for carbonic acid are K₁ = 4.2 x 10^-7 and K₂ = 4.8 x 10^-11 The correct statement for a saturated 0.034 M solution of carbonic acid is

77. The number of solutions of $10 to the power of 2 divided by x space end exponent plus 25 to the power of 1 divided by x space end exponent equals left parenthesis 65 over 8 right parenthesis left parenthesis 50 to the power of 1 divided by x space end exponent right parenthesis$ is

79. If A= $open vertical bar table row 1 0 row cell 1 half end cell 1 end table close vertical bar$ , then A¹⁰⁰ is equal to

83. Water drops fall at regular intervals from a tap which is 5m above the ground. The third drop is leaving the tap at the instant the first drop touches the ground. How far above the ground is the second drop at that instant ? (Take g=10ms^-2)

86. Fundamental period of the function $f left parenthesis x right parenthesis equals sin x plus 2 sin open parentheses x plus straight pi over 6 close parentheses$ is

87. The locus of the points of intersection of the tangent at the extremities of the chords of the ellipse x² + 2y² = 6 which touch the ellipse x² + 4y² = 4 is

88. A particle is oscillating according to the equation $x equals 7 space cos space 0.5 space pi space t$ , where t is in second. The point moves from the position of equilibrium to maximum displacement in time

89. If I = $integral subscript 1 third end subscript superscript 3 1 over x sin open parentheses 1 over x minus x close parentheses d x$ , then I equals