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Measurement NEET Questions — Free MCQ Practice

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Which of the following group have different dimension

A.

Potential difference, EMF, voltage

B.

Pressure, stress, young's modulus

C.

Heat, energy, work-done

D.

Dipole moment, electric flux, electric field

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Explanation

Dipole moment, electric flux, electric field

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Out of following four dimensional quantities, which one quantity is to be called a dimensional constant

A.

Acceleration due to gravity

B.

Surface tension of water

C.

Weight of a standard kilogram mass

D.

The velocity of light in vacuum

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Explanation

The velocity of light in vacuum

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Density of a liquid in CGS system is 0.625 g/cm³. What is its magnitude in SI system

A.

0.625

B.

0.0625

C.

0.00625

D.

625

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Explanation

CGS SI

$N_{1} U_{1} = N_{2} U_{2}$

$N_{1} [M_{1} L_{1}^{- 3}] = N_{2} [M_{2} L_{2}^{- 3}]$

$∴ N_{2} = N_{1} [\frac{M_{1}}{M_{2}}] \times {[\frac{L_{1}}{L_{2}}]}^{- 3}$ $= 0 .625 [\frac{1 g}{1 k g}] \times {[\frac{1 c m}{1 m}]}^{- 3}$

$= 0 .625 \times 10^{- 3} \times 10^{6} = 625$

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The period of oscillation of a simple pendulum is given by $T = 2 π \sqrt{\frac{l}{g}}$ where l is about 100 cm and is known to have 1mm accuracy. The period is about 2s. The time of 100 oscillations is measured by a stop watch of least count 0.1 s. The percentage error in g is

A.

0.1%

B.

1%

C.

0.2%

D.

0.8%

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Explanation

$T = 2 π \sqrt{l / g}$ $\Rightarrow T^{2} = 4 π^{2} l / g$ $\Rightarrow g = \frac{4 π^{2} l}{T^{2}}$

Here % error in l = $\frac{1 m m}{100 c m} \times 100 = \frac{0 .1}{100} \times 100 = 0 .1 %$ and % error in T = $\frac{0 .1}{2 \times 100} \times 100 = 0 .05 %$

$∴$ % error in g = % error in l + 2(% error in T)

$= 0 .1 + 2 \times 0 .05$ = 0.2 %

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The percentage errors in the measurement of mass and speed are 2% and 3% respectively. How much will be the maximum error in the estimation of the kinetic energy obtained by measuring mass and speed

A.

11%

B.

8%

C.

5%

D.

1%

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Explanation

$∴ E = \frac{1}{2} m v^{2}$

$∴$ % Error in K.E.

= % error in mass + 2 × % error in velocity

= 2 + 2 × 3 = 8 %

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The random error in the arithmetic mean of 100 observations is x; then random error in the arithmetic mean of 400 observations would be

A.

4x

B.

$\frac{1}{4} x$

C.

2x

D.

$\frac{1}{2} x$

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Explanation

Suppose we are measuring Time period (T) of a simple pendulum by taking n observations.

So, Time period T = $\frac{T i m e o f n o b s e r v a t i o n s (t)}{N o . o f o b s e r v a t i o n s (n)}$

$E r r o r i n t i m e - p e r i o d \frac{∆ T}{T} = \frac{∆ t}{t} ∆ t (e r r o r i n t i m e) c o r r e s p o n d s t o l e a s t c o u n t o f i n s t r u m e n t (e x - s t o p w a t c h)$

So more is the number of observations, more is the time of measurement (t), and less is the fractional error $\frac{△ T}{T}$

a) 100 observations (n= 100)

So, $\frac{∆ T}{T_{1}} = \frac{∆ t}{t_{1}} = x (l e t)$

b) 400 observations (n=400 )

$\frac{∆ T}{T_{2}} = \frac{∆ t}{t_{2}} = x_{2} (l e t)$

$t_{2} = 4 t_{1} ∴ x_{2} = \frac{x}{4}$

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What is the number of significant figures in 0.310×10³

A.

2

B.

3

C.

4

D.

6

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Explanation

Number of significant figures are 3, because 10³ is decimal multiplier.

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Error in the measurement of radius of a sphere is 1%. The error in the calculated value of its volume is

A.

1%

B.

3%

C.

5%

D.

7%

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Explanation

$∵ V = \frac{4}{3} π r^{3}$

$∴$ % error is volume $= 3 \times %$ error in radius

$= 3 \times 1$ = 3%

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The mean time period of second's pendulum is 2.00s and mean absolute error in the time period is 0.05s. To express maximum estimate of error, the time period should be written as

A.

(2.00 ± 0.01) s

B.

(2.00 + 0.025) s

C.

(2.00 ± 0.05) s

D.

(2.00 ± 0.10) s

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Explanation

Mean time period T = 2.00 sec

& Mean absolute error $= Δ T$ = 0.05 sec.

To express maximum estimate of error, the time period should be written as $(2 .00 \pm 0 .05)$ sec

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A body travels uniformly a distance of (13.8 $\pm$ 0.2) m in a time (4.0 ± 0.3) s. The velocity of the body within error limits is

A.

(3.45 ± 0.2) ms^-1

B.

(3.45 ± 0.3) ms^-1

C.

(3.45 ± 0.4) ms^-1

D.

(3.45 ± 0.5) ms^-1

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Explanation

Here, $S = (13 .8 \pm 0 .2)$ m

and $t = (4 .0 \pm 0 .3)$ sec

Expressing it in percentage error, we have,

$S = 13 .8 \pm \frac{0 .2}{13 .8} \times 100 % = 13 .8 \pm 1 .4 %$

and $t = 4 .0 \pm \frac{0 .3}{4} \times 100 % = 4 \pm 7 .5 %$

$v = \frac{s}{t} M e a n v a l u e o f v = \frac{s}{t} = \frac{13.8}{4} = 3.45 m / s T a k i n g p e r c e n t a g e e r r o r s o n b o t h s i d e s, \frac{∆ v}{v} \times 100 = \frac{△ s}{s} \times 100 + \frac{∆ t}{t} \times 100 = 1.4 + 7.5 = 8.9 % ∴ \frac{∆ v}{v} = \frac{8.9}{100} = 0.089 ∴ ∆ v = v \times 0.089 = 3.45 \times 0.089 = 0.3 m / s R e p o r t i n g v a l u e = v \pm ∆ v = (3 .45 \pm 0 .3) m / s .$

Measurement MCQs for NEET — Physics Questions with Answers

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