﻿ Medical Entrance Exam Question and Answers | Units and Measurement - Zigya

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# Units and Measurement

#### Multiple Choice Questions

1.

The velocity v of a particle at time t is given by where a, b and c are constants, The dimensions of a, b and c are respectively:

• • • • A. According to principle of homogeneity of dimensions, the dimensions of all the terms in a physical expression should be same.
The given expression is From principle of homogeneity 502 Views

2.

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

D.

Force F = at + bt2

From the principle of homogeneity

Dimension of at = dimension of F

$\therefore \mathrm{Dimension}\mathrm{of}\mathrm{a}=\frac{\left[\mathrm{F}\right]}{\left[\mathrm{t}\right]}\phantom{\rule{0ex}{0ex}}=\frac{\left[\mathrm{M}\mathrm{L}{\mathrm{T}}^{-2}\right]}{\left[\mathrm{T}\right]}\phantom{\rule{0ex}{0ex}}=\left[\mathrm{M}\mathrm{L}{\mathrm{T}}^{-3}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Dimension}\mathrm{of}\mathrm{b}=\mathrm{dimension}\mathrm{of}\mathrm{B}\phantom{\rule{0ex}{0ex}}\mathrm{b}=\frac{\left[\mathrm{F}\right]}{\left[{\mathrm{t}}^{2}\right]}=\frac{\left[\mathrm{M}\mathrm{L}{\mathrm{T}}^{-2}\right]}{\left[{\mathrm{T}}^{2}\right]}\phantom{\rule{0ex}{0ex}}=\left[\mathrm{M}\mathrm{L}{\mathrm{T}}^{-4}\right]$

3.

The dimensional formula for Young's modulus is

• [ML-1T-2]

• [M0LT-2]

• [MLT-2]

• [ML2T-2]

A.

[ML-1T-2]

The formula for Young's modulus is

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

${\mathrm{Y}}{}{=}{}\frac{{\mathrm{ML}}^{-1}{\mathrm{T}}^{-2}}{{\mathrm{M}}^{0}{\mathrm{L}}^{0}{\mathrm{T}}^{0}}\phantom{\rule{0ex}{0ex}}{=}{}\left[{\mathrm{ML}}^{-1}{\mathrm{T}}^{-2}\right]$

4.

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

B.

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

5.

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

B.

FAT2

Let energy E = Fx Ay T3

Writing the dimensions on both sides

$\left[\mathrm{M}{\mathrm{L}}^{2}{\mathrm{T}}^{-2}\right]={\left[\mathrm{M} \mathrm{L} {\mathrm{T}}^{-2}\right]}^{\mathrm{x}}{\left[\mathrm{L}{\mathrm{T}}^{-2}\right]}^{\mathrm{y}}\left[\mathrm{T}\right]{}^{2}\phantom{\rule{0ex}{0ex}}\left[\mathrm{M}{\mathrm{L}}^{2}{\mathrm{T}}^{-2}\right]=\left[{\mathrm{M}}^{\mathrm{x}}{\mathrm{L}}^{\mathrm{x}+\mathrm{y}}{\mathrm{T}}^{-2\mathrm{x}-2\mathrm{y}+\mathrm{z}}\right]\phantom{\rule{0ex}{0ex}}\mathrm{On}\mathrm{comparing}\mathrm{both}\mathrm{sides}\phantom{\rule{0ex}{0ex}}\mathrm{x}=1\phantom{\rule{0ex}{0ex}}\mathrm{x}+\mathrm{y}=1\phantom{\rule{0ex}{0ex}}⇒\mathrm{y}=2-\mathrm{x}=2=2-1=1\phantom{\rule{0ex}{0ex}}\mathrm{and}-2\mathrm{x}-2\mathrm{y}+\mathrm{z}=-2\phantom{\rule{0ex}{0ex}}⇒\mathrm{z}=-2×1-2×1+\mathrm{z}\phantom{\rule{0ex}{0ex}}\therefore \mathrm{Dimensional}\mathrm{formula}\mathrm{of}\mathrm{energy}\phantom{\rule{0ex}{0ex}}=\mathbf{F}\mathbf{}\mathbf{A}\mathbf{}{\mathbf{T}}^{\mathbf{2}}$

6.

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]

D.

[ML2T-1I-1]

Resistance 484 Views

7.

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

A.

M0L0T

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

8.

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

B.

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

9.

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

• $\left[{\mathrm{ML}}^{5/2}{\mathrm{T}}^{-2}\right]$

• $\left[{\mathrm{ML}}^{2}{\mathrm{T}}^{-2}\right]$

• $\left[{\mathrm{M}}^{3/2}{\mathrm{L}}^{3/2}{\mathrm{T}}^{-2}\right]$

• $\left[{\mathrm{ML}}^{7/2}{\mathrm{T}}^{-2}\right]$

D.

$\left[{\mathrm{ML}}^{7/2}{\mathrm{T}}^{-2}\right]$

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

10.

The SI unit of gravitational potential is

• J

• J-kg-1

• J-kg

• J-kg2

B.

J-kg-1

$\mathrm{Gravitational}\mathrm{potential}=\frac{\mathrm{work}}{\mathrm{mass}}\phantom{\rule{0ex}{0ex}}\mathrm{Hence},\mathrm{SI}\mathrm{unit}\mathrm{of}\mathrm{gravitational}\mathrm{potential}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{unit}\mathrm{of}\mathrm{work}}{\mathrm{unit}\mathrm{of}\mathrm{mass}}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{J}}{\mathrm{kg}}=\mathrm{J}-{\mathrm{kg}}^{-1}$