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Measurement MCQs for NEET — Physics Questions with Answers

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A physical quantity P is given by P = $\frac{A^{3} B^{\frac{1}{2}}}{C^{- 4} D^{\frac{3}{2}}}$ . The quantity which brings in the maximum percentage error in P is

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Explanation

Quantity C has maximum power. So it brings maximum error in P.

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If L = 2.331 cm, B = 2.1 cm, then L + B =

4.431 cm

4.43 cm

4.4 cm

4 cm

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Explanation

Given, L = 2.331 cm

= 2.33 (correct upto two decimal places)

and B = 2.1 cm = 2.10 cm

$∴ L + B = 2 .33 + 2 .10 = 4 .43 c m$ = 4.4 cm

Since minimum significant figure is 2.

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The number of significant figures in all the given numbers 25.12, 2009, 4.156 and $1 .217 \times 10^{- 4}$ is

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Explanation

The number of significant figures in all of the given number is 4.

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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

$a α + b β + c γ$

$a α + b β - c γ$

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

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

12%

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

360000

72,000

36,000

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]

$\frac{1}{R C}$

$\frac{R}{L}$

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

$\frac{C}{L}$

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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

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

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

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

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

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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

Only numerically correct

Only dimensionally correct

Both numerically and dimensionally correct

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

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

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

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

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

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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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