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Physics NEET Questions — Free MCQ Practice

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

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

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The unit of percentage error is

A.

Same as that of physical quantity

B.

Different from that of physical quantity

C.

Percentage error is unit less

D.

Errors have got their own units which are different from that of physical quantity measured

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Explanation

Percentage error is unit less

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The decimal equivalent of 1/20 upto three significant figures is

A.

0.0500

B.

0.05000

C.

0.0050

D.

5.0 × 10^-2

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Explanation

$\frac{1}{20} = 0 .05$

$∴$ Decimal equivalent upto 3 significant figures is 0.0500

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Accuracy of measurement is determined by

A.

Absolute error

B.

Percentage error

C.

Both

D.

None of these

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Explanation

Percentage error

Physics MCQs for NEET — Practice Questions with Answers

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