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NEET NEET Questions — Free MCQ Practice

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The possible value of Poisson's ratio is

A.

1

B.

0.9

C.

0.8

D.

0.4

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Explanation

(d) Poisson’s ratio varies between – 1 and 0.5.

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Which of the following affects the elasticity of a substance

A.

Hammering and annealing

B.

Change in temperature

C.

Impurity in substance

D.

All of these

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Explanation

(d) The elasticity of substance is affected by impurity of substance hammering and anneasing change in the temperature.

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A wire of diameter 1mm breaks under a tension of 1000 N. Another wire, of same material as that of the first one, but of diameter 2 mm breaks under a tension of

A.

500 N

B.

1000 N

C.

10000 N

D.

4000 N

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Explanation

(d) Breaking force $\propto$ r²

If diameter becomes double then breaking force will become four times i.e. 1000 × 4 = 4000 N

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The force required to stretch a steel wire of cross-section $1 {cm}^{2}$ to 1.1 times its length would be $(Y = 2 \times 10^{11} N m^{- 2})$

A.

$2 \times 10^{6} N$

B.

$2 \times 10^{3} N$

C.

$2 \times 10^{- 6} N$

D.

$2 \times 10^{- 7} N$

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Explanation

(a) $F = A \times Y \times s t r a i n = 1 \times 10^{- 4} \times 2 \times 10^{11} \times 0.1 = 2 \times 10^{6} N$

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When a weight of 10 kg is suspended from a copper wire of length 3 metres and diameter 0.4 mm, its length increases by 2.4 cm. If the diameter of the wire is doubled, then the extension in its length will be

A.

9.6 cm

B.

4.8 cm

C.

1.2 cm

D.

0.6 cm

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Explanation

(d)

$l \propto \frac{1}{r^{2}} (F, L a n d Y a r e c o n s \tan t) \frac{l_{2}}{l_{1}} = {(\frac{r_{1}}{r_{2}})}^{2} = {(\frac{1}{2})}^{2} \Rightarrow l_{2} = \frac{l_{1}}{4} = \frac{2.4}{4} \Rightarrow l_{2} = 0.6 c m$

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A fixed volume of iron is drawn into a wire of length L. The extension x produced in this wire by a constant force F is proportional to

A.

$\frac{1}{L^{2}}$

B.

$\frac{1}{L}$

C.

$L^{2}$

D.

L

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Explanation

(c) $l = \frac{F L}{A Y} = \frac{F L^{2}}{(A L) Y} = \frac{F L^{2}}{V Y}$

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On applying a stress of $20 \times 10^{8}$ N/ $m^{2}$ the length of a perfectly elastic wire is doubled. Its Young’s modulus will be

A.

$40 \times 10^{8} N / m^{2}$

B.

$20 \times 10^{8} N / m^{2}$

C.

$10 \times 10^{8} N / m^{2}$

D.

$5 \times 10^{8} N / m^{2}$

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Explanation

(b) Young's modules = $\frac{s t r e s s}{s t r a i n}$

As the length of wire get doubled therefore strain=1

$∴ Y = s t r a i n = 20 \times 10^{8} N / m^{2}$

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The length of an elastic string is a metre when the longitudinal tension is 4 N and b metre when the longitudinal tension is 5 N. The length of the string in metre when the longitudinal tension is 9 N is

A.

a - b

B.

5b - 4a

C.

2b - $\frac{1}{4} a$

D.

4a - 3b

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Explanation

(b) Let L is the original length of the wire and K is force constant of wire.

Final length = initial length + elongation

$L^{1} = L + \frac{F}{K} F o r f i r s t c o n d i t i o n α = L + \frac{4}{K} . . . . (i) F o r s e c o n d c o n d i t i o n b = L + \frac{5}{K} . . . . (i i) B y s o l v i n g (i) a n d (i i) e q u a t i o n w e g e t L = 5 a - 4 b a n d K = \frac{1}{b - a} N o w w h e n t h e l o n g i t u d i n a l t e n s i o n i s 9 N, l e n g t h o f t h e s t r i n g = L + \frac{9}{K} = 5 a - 4 b + 9 (b - a) = 5 b - 4 a .$

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How much force is required to produce an increase of 0.2% in the length of a brass wire of diameter 0.6 mm ?

(Young’s modulus for brass = $0.9 \times 10^{11} N / m^{2}$ )

A.

Nearly 17 N

B.

Nearly 34 N

C.

Nearly 51 N

D.

Nearly 68 N

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Explanation

(c) $F = \frac{Y A l}{L} = 0.9 \times 10^{11} \times π \times {(0.3 \times 10^{- 3})}^{2} \times \frac{0.2}{100} = 51 N$

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A 5 m long aluminium wire $(Y = 7 \times 10^{10} N / m^{2})$ of diameter 3 mm supports a 40 kg mass. In order to have the same elongation in a copper wire $(Y = 12 \times 10^{10} N / m^{2})$ of the same length under the same weight, the diameter should now be, in mm

A.

1.75

B.

1.5

C.

2.5

D.

5.0

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Explanation

(c)

$l = \frac{F L}{{πr}^{2} Y} \Rightarrow r^{2} \propto \frac{1}{Y} (F, L a n d l a r e c o n s \tan t) \frac{r_{2}}{r_{1}} = {(\frac{Y_{1}}{Y_{2}})}^{1 / 2} = {(\frac{7 \times 10^{10}}{12 \times 10^{10}})}^{1 / 2} \Rightarrow r_{2} = 1.5 \times {(\frac{7}{12})}^{1 / 2} = 1.145 m m ∴ d i a = 2.29 m m$

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