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NEET NEET Questions — Free MCQ Practice

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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

A.

1 × 10² cm³

B.

2 × 10² cm³

C.

1.763 × 10² cm³

D.

None of these

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Explanation

Volume V = l × b × t

$= 12 \times 6 \times 2 .45$ $= 176 .4 c m^{3}$

V $= 1 .764 \times 10^{2} c m^{3}$

since, the minimum number of significant figure is one in breadth, hence volume will also contain only one significant figure. Hence, $V = 2 \times 10^{2} c m^{3}$ .

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A physical quantity of the dimensions of length that can be formed out of c,G and $\frac{e^{2}}{4 π ε_{ο}}$ is [c is velocity of light, G is universal constant of gravitation and e is charge]

A.

$\frac{1}{c^{2}} {[G \frac{e^{2}}{4 π ε_{o}}]}^{\frac{1}{2}}$

B.

$c^{2} {[G \frac{e^{2}}{4 π ε_{o}}]}^{\frac{1}{2}}$

C.

$\frac{1}{c^{2}} {[\frac{e^{2}}{G 4 π ε_{o}}]}^{\frac{1}{2}}$

D.

$\frac{1}{c} G \frac{e^{2}}{4 π ε_{0}}$

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Explanation

(a) As force F= $\frac{e^{2}}{4 π ε_{o} r^{2}}$

$\Rightarrow \frac{e^{2}}{4 π ε_{o}} = r^{2} . F$

putting dimensions of r and F, we get

$\Rightarrow [\frac{e^{2}}{4 π ε_{o}}] = [M L^{3} T^{- 2}] . . . (i)$

Also force. F= $\frac{G m^{2}}{r^{2}}$

$\Rightarrow |G| = \frac{[M L T^{- 2}] [L^{2}]}{[M^{2}]}$ ...(ii)

$\Rightarrow |G| = [M^{- 1} L^{3} T^{- 2}]$ ...(iii)

and $[\frac{1}{c^{2}}] = \frac{1}{[L^{2} T^{- 2}]} = [L^{- 2} T^{2}]$

Now, checking optionwise.

= $\frac{1}{c^{2}} {(\frac{G e^{2}}{4 π ε_{o}})}^{\frac{1}{2}}$

= $[L^{- 2} T^{2}] {[L^{6} T^{- 4}]}^{\frac{1}{2}} = |L|$

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Planck's constant (h),speed of light in vacuum (c) and Newton's gravitational constant (G) are three fundamental constants. Which of the following combinations of these has the dimensions of lenghth?

A.

$\frac{\sqrt{h G}}{c^{3 / 2}}$

B.

$\frac{\sqrt{h G}}{c^{5 / 2}}$

C.

$\sqrt{\frac{h c}{G}}$

D.

$\sqrt{\frac{G c}{h^{3 / 2}}}$

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Explanation

(a) In forms of h,c and G length can be expressed as L= ${(h)}^{a} {(c)}^{b} {(G)}^{c}$

writing dimensions on both sides,we get $[M^{o} L T^{o}] = {[M L^{2} T^{- 1}]}^{a} {[L T^{- 1}]}^{b} {[M^{- 1} L^{3} T^{- 2}]}^{c} = M^{a - c} L^{2 a + b + 3 c} T^{- a - b - 2 c}$

On comparing powers of M,L and T on both sides, we get

$a - c = 0, 2 a + b + 3 c = 1 a n d - a - b - 2 c = 0$

On solving we, get

a=c = $\frac{1}{2} a n d b = - \frac{3}{2}$

$∴ D i m e n s i o n s$ length is

$L = {(h)}^{1 / 2} {(c)}^{- 3 / 2} {(G)}^{1 / 2} = \frac{\sqrt{h G}}{c^{3 / 2}}$

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If energy (E), velocity (v) and time (T) are chosen as the fundamental quantities, the dimensional formula of surface tension will be

A.

$[\begin{matrix} E & v^{- 2} T^{- 1} \end{matrix}]$

B.

[ $E v^{- 1} T^{- 2}$ ]

C.

$[\begin{matrix} E v^{- 2} & T^{- 2} \end{matrix}]$

D.

[ $E^{- 2} v^{- 1_{T} - 3}$ ]

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Explanation

We know that Surface tension(S)=Force[F]/Length[L]
So, [S]=[MLT^-2]/[L]=[ML⁰T^-2]

Energy[E]=Force x displacement
=> [E]=[ML²T²]

Velocity(v)=displacement/time

=>[v]=[LT^-1]

As, S∝E^av^bT^C
where, a,b,c are constants.

From the principle of homogeneity ^[LHS]=[RHS]=>[ML⁰T^-2]= ${[M L^{2} T}^{- 2]^{a}}$ $[L T^{- 1}]$ ^{b ${[T]}^{c}$

=>[ $M L^{0} T^{- 2}$ ]=[ $M^{a} L^{2 a + b} T^{- 2 a - b}$ +c]

Equating the power on both sides, we get}a=1,2a+b=0,b=-2

=>-2a-b+c=-2

=>c=(2a+b)-2=0-2=-2

So [s]=[Ev^-2T^-2]

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If force (F), velocity (v) and time (T) are taken as fundamental units, then the dimensions of mass are

A.

[FvT^-1]

B.

[FvT^-2]

C.

[Fv^-1T^-1]

D.

[Fv^-1T]

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Explanation

We know that
F=ma

=>F=mv/t

=>m=Ft/v

[M]=[F][T]/[v]=[Fv^-1T]

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The density of material in CGS system of units is $4 g / c m^{3} .$ In a system of units in which unit of length is 10cm and unit of mass is 100g, yhen the value of density of material in this system will be-

A.

0.4

B.

40

C.

400

D.

0.04

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Explanation

$D e n s i t y, ρ = \frac{M}{V} ρ = 4 \frac{g m}{c m^{3}} I f u n i t o f m a s s = 100 g m \Rightarrow 1 g m = \frac{1}{100} u n i t o f m a s s A n d, i f u n i t o f l e n g t h = 10 c m \Rightarrow 1 c m = \frac{1}{10} u n i t o f l e n g t h T h e n, D e n s i t y, ρ = \frac{\frac{4}{100} u n i t o f m a s s}{\frac{1}{10^{3}} (u n i t o f l e n g t h)^{3}} = 40 \frac{u n i t o f m a s s}{(u n i t o f l e n g t h)^{3}}$

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The dimension of $\frac{1}{2} ε_{0} E^{2},$ where $ε_{0}$ is permittivity of free space and E is electric field, is

A.

$[M L^{2} T^{- 2}]$

B.

$[M L^{- 1} T^{- 2}]$

C.

$[M L^{2} T^{- 1}]$

D.

$[M L T^{- 1}]$

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Explanation

Using Coulomb's law

$F = \frac{k q_{1} q_{2}}{r^{2}} w h e r e k = \frac{1}{4 π \in_{o}} F = (\frac{1}{4 π \in_{o}}) \frac{q_{1} q_{2}}{r^{2}} S e p a r a t i n g \in_{o}, \in_{o} = (\frac{1}{4 πF}) \frac{q_{1} q_{2}}{r^{2}} N o w d i m e n s i o n s o f \in_{o}, [\in_{o}] = \frac{{(A T)}^{2}}{(M L T^{- 2}) X L^{2}} = M^{- 1} A^{2} L^{- 3} T^{4}$

Dimensions of $ε_{0} = [M^{- 1} L^{- 3} T^{4} A^{2}]$

Dimensions of E= $[M L T^{- 3} A^{- 1}]$

$∴ D i m e n s i o n s o f \frac{1}{2} ε_{0} E^{2} = [M^{- 1} L^{- 3} T^{4} A^{2}] \times [M^{2} L^{2} T^{- 6} A^{- 2}] = [M L^{- 1} T^{- 2}]$

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If the dimension of a physical quantity are given by $M^{a} L^{b} T^{c},$ then the physical quantity will be

A.

pressure if a=1, b=-1, c=-2

B.

velocity if a=1, b=0, c=-1

C.

acceleration if a=1, b=1,c=-2

D.

force if a=0, b=-1, c=-2

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Explanation

(i) Dimensions of velocity $= [M^{0} L^{1} T^{- 1}]$

Here, a=0, b=1, c=-1

(ii) Dimensions of acceleration $= [M^{0} L^{1} T^{- 2}]$

Here, a=0, b=1, c=-2

(iii) Dimensions of force $= [M^{1} L^{1} T^{- 2}]$

Here, a=1, b=1, T=-2

(iv) Dimensions of pressure $= [M^{1} L^{- 1} T^{- 2}]$

$∴$ Here, a=1, b=-1, c=-2

$∴$ The physical quantity is pressure.

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Which two of the following five physical parameters have the same dimensions ?

A.

energy density

B.

refractive index

C.

dielectric constant

D.

Young's modulus

(5) magnetic field

(a) 2 and 4

(b) 3 and 5

(c) 1 and 4

(d) 1 and 5

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Explanation

Energy density= $\frac{E n e r g y}{V o l u m e} = \frac{[M L^{2} T^{- 2}]}{[L^{3}]} = [M L^{- 1} T^{- 2}]$

Refractive index hole no dimensions.

Dielectric constant has no dimensions.

Young's modulus,

$Y = \frac{F l}{A ∆ l} = \frac{[M L T^{- 3}] [L]}{[L^{2}] [L]} = [M L^{- 1} T^{- 2}]$

(5) Magnetic field,

$B = \frac{F}{I l} = \frac{[M L T^{- 2}]}{[A] [L]} = [M T^{- 2} A^{- 1}]$

Therefore,option (c) is correct

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In a vernier calliper N divisions of vernier scale coincides with N-1 divisions of main scale (in which length of one division is 1 mm). The least count of the instrument should be

A.

$N mm$

B.

$N - 1 mm$

C.

$\frac{1}{10 N} cm$

D.

$\frac{1}{N - 1} mm$

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Explanation

Least Count = 1MSD - 1VSD

$N VSD = (N-1) MSD 1 V S D = \frac{N - 1}{N} M S D L C = 1 M S D - \frac{N - 1}{N} M S D = \frac{1}{N} M S D = \frac{1}{10 N} c m$

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