Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

Units and Measurement

Multiple Choice Questions

The velocity v of a particle at time t is given by $straight v equals at plus fraction numerator straight b over denominator straight t plus straight c end fraction comma$ where a, b and c are constants, The dimensions of a, b and c are respectively:

$open square brackets LT to the power of negative 2 end exponent close square brackets comma space open square brackets straight L close square brackets space and space open square brackets straight T close square brackets$
$open square brackets straight L squared close square brackets space open square brackets straight T close square brackets space and space open square brackets LT squared close square brackets$
$open square brackets LT squared close square brackets comma space open square brackets LT close square brackets space and space open square brackets straight L close square brackets$
$open square brackets LT squared close square brackets comma space open square brackets LT close square brackets space and space open square brackets straight L close square brackets$

open square brackets LT to the power of negative 2 end exponent close square brackets comma space open square brackets straight L close square brackets space and space open square brackets straight T close square brackets

According to principle of homogeneity of dimensions, the dimensions of all the terms in a physical expression should be same.
The given expression is
$straight v equals at plus fraction numerator straight b over denominator straight t plus straight c end fraction$
From principle of homogeneity
$open square brackets straight a close square brackets space open square brackets straight t close square brackets space equals space open square brackets straight v close square brackets space space space space space open square brackets straight a close square brackets space equals space fraction numerator open square brackets straight v close square brackets over denominator open square brackets straight t close square brackets end fraction space equals space fraction numerator open square brackets LT to the power of negative 1 end exponent close square brackets over denominator open square brackets straight T close square brackets end fraction space equals space open square brackets LT to the power of negative 2 end exponent close square brackets Similarly comma space space open square brackets straight c close square brackets space equals space open square brackets straight t close square brackets space equals space open square brackets straight T close square brackets Further comma space space space space space fraction numerator open square brackets straight b close square brackets over denominator open square brackets straight t plus straight c close square brackets end fraction space equals open square brackets straight v close square brackets or space space space space space space space space space space space space open square brackets straight b close square brackets space equals space open square brackets straight v close square brackets space open square brackets straight t plus straight c close square brackets or space space space space space space space space space space space space open square brackets straight b close square brackets space equals space open square brackets LT to the power of negative 1 end exponent close square brackets space open square brackets straight T close square brackets space equals space open square brackets straight L close square brackets$

502 Views

A force F is given by F =at + bt² ,where, t istime. What are the dimensions of a and b ?

$[M L T^{- 1}] and [M L T^{0}]$
$[M L T^{- 3}] and [M L^{2} T^{4}]$
$[M L T^{- 4}] and [M L T^{1}]$
$[M L T^{- 3}] and [M L T^{- 4}]$

$[M L T^{- 3}] and [M L T^{- 4}]$

Force F = at + bt²

From the principle of homogeneity

Dimension of at = dimension of F

$∴ Dimension of a = \frac{[F]}{[t]} = \frac{[M L T^{- 2}]}{[T]} = [M L T^{- 3}] Dimension of b = dimension of B b = \frac{[F]}{[t^{2}]} = \frac{[M L T^{- 2}]}{[T^{2}]} = [M L T^{- 4}]$

The dimensional formula for Young's modulus is

[ML^-1T^-2]
[M⁰LT^-2]
[MLT^-2]
[ML²T^-2]

[ML^-1T^-2]

The formula for Young's modulus is

Y = $\frac{stress}{strain} = \frac{F / A}{Δl / l}$

$Y = \frac{{ML}^{- 1} T^{- 2}}{M^{0} L^{0} T^{0}} = [{ML}^{- 1} T^{- 2}]$

The length, breadth and thickness of ablock are given by l = 12 cm, b = 6 cm andt = 2.45 cm. The volume of the block according to the idea of significant figures should be

1 × 10² cm³
2 × 10² cm³
1.76 × 10² cm³
None of these

2 × 10² cm³

Using relation for volume

Given:- length of a block = 12 cm

breadth of a block = 6 cm

thickness of a block = 2.45cm

V =length × breadth × thickness

=12 × 6 × 2.45 = 176.4

=1.764 ×10² cm³

The minimum number of significant figures is 1in thickness, hence the volume will contain only onesignificant figure.

Therefore V= 2 ×10² cm³

In a system of units if force (F), acceleration (A),and time (T) are taken as fundamental units,then the dimensional formula of energy is:

FA²T
FAT²
F²AT
FAT

FAT²

Let energy E = F^x A^y T³

Writing the dimensions on both sides

$[M L^{2} T^{- 2}] = {[M L T^{- 2}]}^{x} {[L T^{- 2}]}^{y} [T]^{2} [M L^{2} T^{- 2}] = [M^{x} L^{x + y} T^{- 2 x - 2 y + z}] On comparing both sides x = 1 x + y = 1 \Rightarrow y = 2 - x = 2 = 2 - 1 = 1 and - 2 x - 2 y + z = - 2 \Rightarrow z = - 2 \times 1 - 2 \times 1 + z ∴ Dimensional formula of energy = F A T^{2}$

Dimensions of resistance in an electrical circuit, in terms of the dimension of mass M, of length L, of time T and of current I, would be:

[ML²T^-3I^-1]
[ML²T^-2]
[ML²T^-1I^-1]
[ML²T^-1I^-1]

[ML²T^-1I^-1]

Resistance

$straight R space equals space fraction numerator potential space difference over denominator current end fraction space equals space straight V over straight i space equals space straight W over qi left parenthesis therefore space potnetial space difference space is space equal space to space work space done space per space unit space charge right parenthesis equals space fraction numerator left square bracket Dimensions space of space work right square bracket over denominator left square bracket Dimensions space of space charge right square bracket left square bracket Dimensions space of space current right square bracket end fraction space equals space fraction numerator left square bracket ML squared straight T to the power of negative 2 end exponent right square bracket over denominator left square bracket IT right square bracket left square bracket straight I right square bracket end fraction space equals space left square bracket ML squared straight T to the power of negative 3 end exponent straight I to the power of negative 2 end exponent right square bracket$

484 Views

If C and R denote capacity and resistance, the dimensions of C R are :

M⁰L⁰T
ML⁰T
M⁰L⁰T²
not expressible in terms of M, L and T

M⁰L⁰T

RC is the time constant of the circuit, so its dimensions are [M⁰L⁰T].

The length, breadth and thickness of a block are given by l = 12 cm, b = 6 cm and t = 2.45 cm. The volume of the block according to the idea of significant figures should be:

1 × 10² cm³
2 × 10² cm³
1.763 × 10² cm³
none of the above

2 × 10² cm³

Using the relation for volume
V = length × breadth × thickness
= 12 × 6 × 2.45
= 176.4 cm
= 1.764 × 10² cm³

The minimum number of significant figures is 1 in breadth, hence, the volume will contain only one significant figure. Therefore,

V = 2× 10² cm³

The potential energy of a particle varies with distance x from a fixed origin as $V = (\frac{A \sqrt{x}}{x + B})$ ;where A and B are constants. The dimensions of AB are

$[{ML}^{5 / 2} T^{- 2}]$
$[{ML}^{2} T^{- 2}]$
$[M^{3 / 2} L^{3 / 2} T^{- 2}]$
$[{ML}^{7 / 2} T^{- 2}]$

$[{ML}^{7 / 2} T^{- 2}]$

$Given, v = \frac{A \sqrt{x}}{x + B} . . . . (i) Dimensions of v = dimensions of potential energy = [{ML}^{2} T^{- 2}] From Eq . (i), Dimensions of B = dimensions ofx = [M^{0} {LT}^{0}] ∴ Dimensions of A = \frac{dimensions of v \times dimension of (x + B)}{dimensions of \sqrt{x}} = \frac{[{ML}^{2} T^{- 2}] [M^{0} {LT}^{0}]}{[M^{0} L^{1 / 2} T^{0}]} = [{ML}^{5 / 2} T^{- 2}] Hence, dimensions of AB = [{ML}^{5 / 2} T^{- 2}] [M^{0} {LT}^{0}] = [{ML}^{7 / 2} T^{- 2}]$

10.

The SI unit of gravitational potential is

J
J-kg^-1
J-kg
J-kg²

J-kg^-1

$Gravitational potential = \frac{work}{mass} Hence, SI unit of gravitational potential = \frac{unit of work}{unit of mass} = \frac{J}{kg} = J - {kg}^{- 1}$