# JEE Main: Units, Dimensions And Error Analysis! Quiz

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Tanmay Shankar
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• 1.

### Suppose refractive index is given as, where P and Q are constants and v is the speed of light and  is the wavelength, then dimensions of Q are the same as that of

• A.

Wavelength

• B.

Volume

• C.

Force

• D.

Acceleration

D. Acceleration
• 2.

### The Equation of stationary wave is then which of the following is false:

D.
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].

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• 3.

### In the Equation   the dimensional formula for will be

• A.

MLT

• B.
• C.
• D.
C.
Explanation
The trigonometrical ratios are the dimensionless quantities so for the equation   the term  will be dimensionless

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• 4.

### For any spring of spring constant k, the frequency of vibrations ‘f’ is given by the formula f = Xmakb, where X is a dimensionless constant. The value of ‘a’ and ‘b’ are, respectively:

D.
Explanation
f = Xmakb 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.

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• 5.

### 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?

• A.

C/g

• B.

Pg

• C.

PCg

• D.

C/P

D. C/P
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 = CagbPc; [M] = [LT-1] a [LT-2] b [ml-1T-2] c,
[M1] = [La+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.

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• 6.

### 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:

• A.

6.5

• B.

4.3

• C.

5.8

• D.

4.9

D. 4.9
Explanation
In Searle’s Experiment, Young’s Modulus Y =  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 = so the relative error in Y;

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• 7.

### A physical Quantity V depends on time as V = V0 (1+ e), where is a constant and t is time, then which of the following is true:

• A.

Î± is a dimensionless constant

• B.

Has dimensions of T2

• C.

Î± has dimensions of V

• D.

Has dimensions of LT-2

B. Has dimensions of T2
Explanation
In the expression V = V0 (1+ ), the term 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-2 i.e. equal to the dimensions of T2.

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• 8.

### The banking of roads is explained with the formula  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.

• A.

Above expression may be dimensionally correct but numerically it is wrong

• B.

Above expression is dimensionally as well as numerically correct

• C.

Above expression is dimensionally as well as numerically wrong

• D.

Above expression is dimensionally incorrect but numerically it is correct

B. Above expression is dimensionally as well as numerically correct
Explanation
The formula is tan dimensionally correct as both the sides are dimensionless. L.H.S. i.e. is a dimensionless trigonometrical ratio and  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.

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• 9.

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

A.
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]

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• 10.

### What will be the dimensions a b of according to the relation where E =  is Energy, x is displacement and t is time?

C.
Explanation
In the equation E =  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].

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• Mar 19, 2023
Quiz Edited by
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• Jul 31, 2014
Quiz Created by
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