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Physics NEET Questions — Free MCQ Practice

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A physical quantity is given by $X = M^{a} L^{b} T^{c}$ . The percentage error in measurement of M, L and T are $α, β$ and $γ$ respectively. Then maximum percentage error in the quantity X is

A.

$a α + b β + c γ$

B.

$a α + b β - c γ$

C.

$\frac{a}{α} + \frac{b}{β} + \frac{c}{γ}$

D.

None of these

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Explanation

Percentage error in X $= a α + b β + c γ$

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A physical quantity A is related to four observable a, b, c and d as follows, $A = \frac{a^{2} b^{3}}{c \sqrt{d}}$ , the percentage errors of measurement in a, b, c and d are 1%,3%,2% and 2% respectively. What is the percentage error in the quantity A

A.

12%

B.

7%

C.

5%

D.

14%

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Explanation

Percentage error in A $= (2 \times 1 + 3 \times 3 + 1 \times 2 + \frac{1}{2} \times 2) % = 14 %$

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If the acceleration due to gravity is 10 ms^–2 and the units of length and time are changed in kilometer and hour respectively, the numerical value of the acceleration is

A.

360000

B.

72,000

C.

36,000

D.

129600

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Explanation

$n_{2} = n_{1} {[\frac{L_{1}}{L_{2}}]}^{1} {[\frac{T_{1}}{T_{2}}]}^{- 2}$ $= 10 {[\frac{m e t e r}{k m}]}^{1} {[\frac{\sec}{h r}]}^{- 2}$

$n_{2} = 10 {[\frac{m}{10^{3} m}]}^{1} {[\frac{\sec}{3600 \sec}]}^{- 2}$ = 129600

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If L, C and R represent inductance, capacitance and resistance respectively, then which of the following does not represent dimensions of frequency [Only for droppers]

A.

$\frac{1}{R C}$

B.

$\frac{R}{L}$

C.

$\frac{1}{\sqrt{L C}}$

D.

$\frac{C}{L}$

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Explanation

$f = \frac{1}{2 π \sqrt{L C}}$ ∴ $(\frac{C}{L})$ does not represent the dimension of frequency.

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Number of particles is given by $n = - D \frac{n_{2} - n_{1}}{x_{2} - x_{1}}$ crossing a unit area perpendicular to X-axis in unit time, where n₁ and n₂ are number of particles per unit volume for the value of x meant to x₂ and x₁. Find dimensions of D called as diffusion constant

A.

$M^{0} L T^{2}$

B.

$M^{0} L^{2} T^{- 4}$

C.

$M^{0} L T^{- 3}$

D.

$M^{0} L^{2} T^{- 1}$

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Explanation

[n] = Number of particles crossing a unit area in unit time = $[L^{- 2} T^{- 1}]$

$[n_{2}] = [n_{1}] =$ number of particles per unit volume = [L^–3]

$[x_{2}] = [x_{1}]$ = positions

∴ $D = \frac{[n] [x_{2} - x_{1}]}{[n_{2} - n_{1}]} = \frac{[L^{- 2} T^{- 1}] \times [L]}{[L^{- 3}]}$ = $[L^{2} T^{- 1}]$

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With the usual notations, the following equation $S_{t} = u + \frac{1}{2} a (2 t - 1)$ is

A.

Only numerically correct

B.

Only dimensionally correct

C.

Both numerically and dimensionally correct

D.

Neither numerically nor dimensionally correct

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Explanation

We can derive this equation from equations of motion so it is numerically correct.

$S_{t}$ = distance travelled in t^th second = $\frac{Distance}{time} = [L T^{- 1}]$

u = velocity = $[L T^{- 1}]$ and $\frac{1}{2} a (2 t - 1) = [L T^{- 1}]$

As dimensions of each term in the given equation are same, hence equation is dimensionally correct also.

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If the dimensions of length are expressed as $G^{x} c^{y} h^{z}$ ; where G, c and h are the universal gravitational constant, speed of light and Planck's constant respectively, then

A.

$x = \frac{1}{2}, y = \frac{1}{2}$

B.

$x = \frac{1}{2}, z = \frac{1}{2}$

C.

$y = \frac{1}{2}, z = \frac{3}{2}$

D.

$y = - \frac{3}{2}, z = \frac{1}{2}$

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Explanation

(2, 4) Length $\propto$ G^xc^yh^z

L= ${[M^{- 1} L^{3} T^{- 2}]}^{x}$ ${[L T^{- 1}]}^{y} {[M L^{2} T^{- 1}]}^{z}$

By comparing the power of M, L and T in both sides we get $- x + z = 0$ , $3 x + y + 2 z = 1$ and $- 2 x - y - z = 0$

By solving above three equations we get

$x = \frac{1}{2}, y = - \frac{3}{2}, z = \frac{1}{2}$

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A highly rigid cubical block A of small mass M and side L is fixed rigidly onto another cubical block B of the same dimensions and of low modulus of rigidity $η$ such that the lower face of A completely covers the upper face of B. The lower face of B is rigidly held on a horizontal surface. A small force F is applied perpendicular to one of the side faces of A. After the force is withdrawn block A executes small oscillations. The time period of which is given by

A.

$2 π \sqrt{\frac{M η}{L}}$

B.

$2 π \sqrt{\frac{L}{M η}}$

C.

$2 π \sqrt{\frac{M L}{η}}$

D.

$2 π \sqrt{\frac{M}{η L}}$

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Explanation

Using dimensional analysis

By substituting the dimensions of mass [M], length [L] and coefficient of rigidity $η$ as $[M L^{- 1} T^{- 2}]$ we get $T = 2 π \sqrt{\frac{M}{η L}}$ is the right formula for time period of oscillations

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The pair(s) of physical quantities that have the do not have same dimensions, is (are)

A.

Reynolds number and coefficient of friction

B.

Latent heat and gravitational potential

C.

Curie and frequency of a light wave

D.

Planck's constant and torque

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Explanation

Pairs (1, 2, 3) have same dimensions

Reynolds number and coefficient of friction are dimensionless.

Latent heat and gravitational potential both have dimension $[L^{2} T^{- 2}]$ .

Curie and frequency of a light wave both have dimension $[T^{- 1}]$ . But dimensions of Planck's constant h is $[M L^{2} T^{- 1}]$ and torque is $[M L^{2} T^{- 2}]$

Energy = h X frequency

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The speed of light (c), gravitational constant (G) and Planck's constant (h) are taken as the fundamental units in a system. The dimension of time in this new system should be

A.

$G^{1 / 2} h^{1 / 2} c^{- 5 / 2}$

B.

$G^{- 1 / 2} h^{1 / 2} c^{1 / 2}$

C.

$G^{1 / 2} h^{1 / 2} c^{- 3 / 2}$

D.

$G^{1 / 2} h^{1 / 2} c^{1 / 2}$

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Explanation

Time $\propto c^{x} G^{y} h^{z} \Rightarrow T = k c^{x} G^{y} h^{z}$

Putting the dimensions in the above relation

$\Rightarrow$ $[M^{0} L^{0} T^{1}] = {[L T^{- 1}]}^{x} {[M^{- 1} L^{3} T^{- 2}]}^{y} {[M L^{2} T^{- 1}]}^{z}$

$\Rightarrow [M^{0} L^{0} T^{1}] = [M^{- y + z} L^{x + 3 y + 2 z} T^{- x - 2 y - z}]$

Comparing the powers of M, L and T

$- y + z = 0$ …(i)

$x + 3 y + 2 z = 0$ …(ii)

$- x - 2 y - z = 1$ …(iii)

On solving equations (i) and (ii) and (iii)

$x = \frac{- 5}{2}, y = z = \frac{1}{2}$

Hence dimension of time are $[G^{1 / 2} h^{1 / 2} c^{- 5 / 2}]$

Physics MCQs for NEET — Practice Questions with Answers

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