JEE Main: Units, Dimensions And Error Analysis! Quiz

1. Suppose refractive index is given as,

mu equals P space plus space fraction numerator Q v squared over denominator lambda end fraction

mu equals P space plus space fraction numerator Q v squared over denominator lambda end fraction

where P and Q are constants and v is the speed of light and

lambda

is the wavelength, then dimensions of Q are the same as that of

Wavelength

Volume

Force

Acceleration

Explanation

2. In the Equation

fraction numerator v t over denominator lambda end fraction equals space 2 A omega S i n left parenthesis omega t plus x subscript 0 right parenthesis

the dimensional formula for will be

MLT

The trigonometrical ratios are the dimensionless quantities so for the equation the term $omega t space plus x subscript 0$ will be dimensionless

Explanation

The trigonometrical ratios are the dimensionless quantities so for the equation the term $omega t space plus x subscript 0$ will be dimensionless

3. The banking of roads is explained with the formula

tan theta space equals space fraction numerator v squared over denominator r g end fraction comma

here v is the speed of the object, r is the radius of the curve formed and g is the gravitational acceleration. Choose the correct statement about this relation.

Above expression may be dimensionally correct but numerically it is wrong

Above expression is dimensionally as well as numerically correct

Above expression is dimensionally as well as numerically wrong

Above expression is dimensionally incorrect but numerically it is correct

The formula is tan $theta space equals space fraction numerator v squared over denominator r g end fraction$ dimensionally correct as both the sides are dimensionless. L.H.S. i.e. is a dimensionless trigonometrical ratio and $fraction numerator v squared over denominator r g end fraction$ is also dimensionless as the dimension of v² and rg are same. Moreover this formula is proved with the help of various theorems and principals so it is also numerically correct.

Explanation

The formula is tan $theta space equals space fraction numerator v squared over denominator r g end fraction$ dimensionally correct as both the sides are dimensionless. L.H.S. i.e. is a dimensionless trigonometrical ratio and $fraction numerator v squared over denominator r g end fraction$ is also dimensionless as the dimension of v² and rg are same. Moreover this formula is proved with the help of various theorems and principals so it is also numerically correct.

4. A physical Quantity V depends on time as V = V₀(1+ e

blank to the power of a t minus 2 end exponent

), where is a constant and t is time, then which of the following is true:

α is a dimensionless constant

Has dimensions of T2

α has dimensions of V

Has dimensions of LT-2

In the expression V = V₀(1+ $e to the power of a t minus 2 end exponent$ ), the term $a t to the power of minus 2 end exponent$ has to be a dimensionless term as it is in the power of an exponential function so for being the dimensionless the dimensions of should be equal to the inverse of the dimensions of T^-2i.e. equal to the dimensions of T².

Explanation

In the expression V = V₀(1+ $e to the power of a t minus 2 end exponent$ ), the term $a t to the power of minus 2 end exponent$ has to be a dimensionless term as it is in the power of an exponential function so for being the dimensionless the dimensions of should be equal to the inverse of the dimensions of T^-2i.e. equal to the dimensions of T².

5. The Equation of stationary wave is

y equals space 2 A space sin left parenthesis fraction numerator 2 πv over denominator lambda end fraction t right parenthesis space cos space left parenthesis fraction numerator 2 pi over denominator lambda end fraction x right parenthesis

then which of the following is false:

All the options must be checked to find out the wrong statement. First option is correct as and are dimensionless quantities. In second option x has the same units as that of i. e. meter. In third option has dimension [L^-1T] which is the inverse of v [LT^-1]. Now only fourth option is left which is incorrect as has the dimension [T^-1], whereas has the dimension [L].

Explanation

All the options must be checked to find out the wrong statement. First option is correct as and are dimensionless quantities. In second option x has the same units as that of i. e. meter. In third option has dimension [L^-1T] which is the inverse of v [LT^-1]. Now only fourth option is left which is incorrect as has the dimension [T^-1], whereas has the dimension [L].

6. What will be the dimensions a

cross times

b of according to the relation where E =

fraction numerator a plus t squared over denominator b x end fraction

is Energy, x is displacement and t is time?

In the equation E = $fraction numerator a plus t squared over denominator b x end fraction$ the dimension of r.h.s. should be equal to the dimension of l.h.s. both sides must have the dimensions equal to the [ML²T^-2]. Now as ‘a’ is added to t² so the dimensions of ‘a’ will be equal to the dimensions of [T²]. For the dimension of ‘b’ [ML²T^-2] = so from here b = [M^-1L^-3T⁴], now a b = [T²] [M^-1L^-3T⁴] = [M^-1L^-3T⁶].

Explanation

In the equation E = $fraction numerator a plus t squared over denominator b x end fraction$ the dimension of r.h.s. should be equal to the dimension of l.h.s. both sides must have the dimensions equal to the [ML²T^-2]. Now as ‘a’ is added to t² so the dimensions of ‘a’ will be equal to the dimensions of [T²]. For the dimension of ‘b’ [ML²T^-2] = so from here b = [M^-1L^-3T⁴], now a b = [T²] [M^-1L^-3T⁴] = [M^-1L^-3T⁶].

7. What will be the dimensions of the mass, if C (velocity of light), g (acceleration due to gravity), and P (atmospheric pressure) are considered as the fundamental Quantities in the MKS system?

C/g

Pg

PCg

C/P

Dimensional formula of velocity of light (C) = [LT^-1]

Dimensional formula of acceleration due to gravity (g) = [LT^-2]

Dimensional formula of atmospheric pressure (P) = [ml^-1T^-2],

Now let M = C^ag^bP^c; [M] = [LT^-1]^a [LT^-2]^b [ml^-1T^-2]^c,

[M¹] = [L^a+b-c] [T^-a-2b-2c] [M ^c];

a+b-c = 0; -a-2b-2c = 0; c = 1; On solving the equations we get, a = 0, b = 1, c = 1,

so, M = Pg.

Explanation

Dimensional formula of velocity of light (C) = [LT^-1]

Dimensional formula of acceleration due to gravity (g) = [LT^-2]

Dimensional formula of atmospheric pressure (P) = [ml^-1T^-2],

Now let M = C^ag^bP^c; [M] = [LT^-1]^a [LT^-2]^b [ml^-1T^-2]^c,

[M¹] = [L^a+b-c] [T^-a-2b-2c] [M ^c];

a+b-c = 0; -a-2b-2c = 0; c = 1; On solving the equations we get, a = 0, b = 1, c = 1,

so, M = Pg.

8. For any spring of spring constant k, the frequency of vibrations 'f' is given by the formula f = Xm^ak^b, where X is a dimensionless constant. The value of 'a' and 'b' are, respectively:

f = Xm^ak^b dimension of f = [T^-1], Dimensions of m = [M], Dimensions of spring constant ‘k’ = [MT^-2] and X is a dimensionless constant.
[T^-1] = [M]^a [MT^-2]^b; [T^-1] = [M]^a+b[T]^-2b, now comparing the powers on both sides,
a+1 = 0; -2b = -1 so b=1/2 and a= -1/2.

Explanation

f = Xm^ak^b dimension of f = [T^-1], Dimensions of m = [M], Dimensions of spring constant ‘k’ = [MT^-2] and X is a dimensionless constant.
[T^-1] = [M]^a [MT^-2]^b; [T^-1] = [M]^a+b[T]^-2b, now comparing the powers on both sides,
a+1 = 0; -2b = -1 so b=1/2 and a= -1/2.

9. The product of permittivity of free space and the permeability of free space has the dimensions equal to the:

Dimension of permittivity of free space ( ) = [M^-1L^-3T⁴A²]
Dimension of permeability of free space ( ) = [MLT^-2A^-2]
Product of and = [M^-1L^-3T⁴A²] [MLT^-2A^-2]
= [L^-2T²]

Explanation

Dimension of permittivity of free space ( ) = [M^-1L^-3T⁴A²]
Dimension of permeability of free space ( ) = [MLT^-2A^-2]
Product of and = [M^-1L^-3T⁴A²] [MLT^-2A^-2]
= [L^-2T²]

10. In Searle's experiment to find Young's modulus, the diameter of the wire is measured as d=0.050 cm, and the length of the wire is L = 100 cm. When a weight of 10.0 kg is placed, the extension in the wire was found to be 0.200 cm. The maximum permissible error (percentage) in Young's modulus is:

6.5

4.3

5.8

4.9

In Searle’s Experiment, Young’s Modulus Y = $fraction numerator F L over denominator A increment L end fraction$ where F is force, A is area, L is the original length and is the change in length due to force F. Here F = Mg and A = 4πd² so Y = $fraction numerator M g L over denominator 4 πd squared space increment straight L end fraction$ so the relative error in Y;

Explanation

In Searle’s Experiment, Young’s Modulus Y = $fraction numerator F L over denominator A increment L end fraction$ where F is force, A is area, L is the original length and is the change in length due to force F. Here F = Mg and A = 4πd² so Y = $fraction numerator M g L over denominator 4 πd squared space increment straight L end fraction$ so the relative error in Y;