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Mechanical Properties of Solids NEET Questions — Free MCQ Practice

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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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A steel wire of 1 m long and cross section area $1 {mm}^{2}$ is hang from rigid end. When mass of 1kg is hung from it then change in length will be (given $Y = 2 \times 10^{11} N / m^{2}$ )

A.

0.5 mm

B.

0.25 mm

C.

0.05 mm

D.

5 mm

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Explanation

(c) $l = \frac{M g L}{Y A} = \frac{1 \times 10 \times 1}{2 \times 10^{11} \times 10^{- 6}} = 0.05 m m$

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A force F is applied on the wire of radius r and length L and change in the length of wire is l. If the same force F is applied on the wire of the same material and radius 2r and length 2L, Then the change in length of the other wire is

A.

l

B.

2l

C.

$\frac{l}{2}$

D.

4l

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Explanation

(c)

$l = \frac{F L}{A Y} \Rightarrow l \propto \frac{L}{r^{2}} (F a n d Y a r e c o n s \tan t) \frac{l_{2}}{l_{1}} = \frac{L_{2}}{L_{1}} \times {(\frac{r_{1}}{r_{2}})}^{2} = 2 \times {(\frac{1}{2})}^{2} = \frac{1}{2} ∴ l_{2} = \frac{l_{1}}{2}$

i.e., the change in the length of other wire is $\frac{l}{2}$

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An iron rod of length 2m and cross section area of 50 X $10^{- 6} m^{2}$ , is stretched by 0.5 mm, when a mass of 250 kg is hung from its lower end. Young's modulus of the iron rod is-

A.

$19.6 \times 10^{10} N / m^{2}$

B.

$19.6 \times 10^{15} N / m^{2}$

C.

$19.6 \times 10^{18} N / m^{2}$

D.

$19.6 \times 10^{20} N / m^{2}$

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Explanation

(a) $Y = \frac{M g L}{A l} = \frac{250 \times 9.8 \times 2}{50 \times 10^{- 6} \times 0.5 \times 10^{- 3}} = 19.6 \times 10^{10} N / m^{2}$

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In which case there is maximum extension in the wire, if same force is applied on each wire

A.

L = 500 cm, d = 0.05 mm

B.

L = 200 cm, d = 0.02 mm

C.

L = 300 cm, d = 0.03 mm

D.

L = 400 cm, d = 0.01 mm

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Explanation

(d) $l \propto \frac{L}{r^{2}} (Y a n d F a r e c o n s \tan t)$

Maximum extension takes place in that wire for which the ratio of $\frac{L}{r^{2}}$ will be maximum.

Mechanical Properties of Solids MCQs for NEET — Physics Questions with Answers

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