Mechanical Properties Of Solids(12th Grade > Physics ) Questions and answers for exam Preparation

12th Grade > Physics

MECHANICAL PROPERTIES OF SOLIDS MCQs

Total Questions : 40 | Page 3 of 4 pages

Question 21. A rubber cord catapult has cross-sectional area

25 m m^{2}

and initial length of rubber cord is 10 cm It is stretched to 5cm and then released to project a missile of mass 5gm.Taking Y_rubber

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

, velocity of projected missile is

20 ms−1
100 ms−1
250 ms−1
200 ms−1

Discuss Question

Answer: Option C. -> 250 ms−1
:
C
c): Potential energy stored in the rubber cord catapult will be converted into kinetic energy of mass.

\frac{1}{2} m v^{2} = \frac{1}{2} \frac{Y A l^{2}}{L} v = \sqrt{\frac{Y A l^{2}}{m L}}

\sqrt{\frac{5 \times 10^{8} \times 25 \times 10^{- 6} \times (5 \times 10^{- 2})^{2}}{5 \times 10^{- 3} \times 10 \times 10^{- 2}}} = 250 m / s

Question 22. Two wires are made of the same material and have the same volume. However, wire 1 has cross sectional area 3A. If length of wire 1 increased by x on applying force F, how much force is needed to stretch wire 2 by the same amount?

Discuss Question

Answer: Option C. -> 9F
:
C

Y = \frac{F l}{A △ l}

so,

F = \frac{Y A^{2} △ l}{A l} = \frac{Y A^{2} △ l}{V}

where,

A l = V =

Volume of wire
hence

F \propto A^{2}

\frac{F^{'}}{F} = \frac{(3 A)^{2}}{A^{2}} = 9

or F'=9F

Question 23. 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 - b
5b - 4a
2b−14a
4a - 3b

Discuss Question

Answer: Option B. -> 5b - 4a
:
B
b) Let L is the original length of the wire and K is force constant of wire.
Final length=initial length+elongation
L' = L +

\frac{F}{K}

For first condition a = L +

\frac{4}{K} . . . . . (i)

For second condition b = L +

\frac{5}{K} . . . . . (i i)

By solving (i) and (ii) equation we get
L = 5a - 4b and K =

\frac{1}{b - a}

Now when the longitudinal tension is 9N, length of the string =L +

\frac{9}{K}

= 5a - 4b + 9(b-a) = 5b-4a.

Question 24. When the tension in a metal wire is

T_{1}

, its length is

l_{1}

. When the tension is

T_{2}

, its length is

l_{2}

. The natural length of wire is

T2T1(l2+l2)
T1l1+T2l2
l1T2−l2T1T2−T1
l1T2+l2T1T2+T1

Discuss Question

Answer: Option C. -> l1T2−l2T1T2−T1
:
C

Y = \frac{F l}{A △ l}

Y, l and A are constants
Therefore

\frac{F}{△ l} = c o s t a n t o r △ l \propto F

Now,

l_{1} - l \propto T_{1} a n d l_{2} - l \propto T_{2}

Dividing,

\frac{l_{1} - l}{l_{2} - l} = \frac{T_{1}}{T_{2}}

l_{1} T_{2} - l T_{2} = l_{2} T_{1} - l T_{1}

l (T_{1} - T_{2}) = l_{2} T_{1} - l_{1} T_{2}

l = \frac{l_{2} T_{1} - l_{1} T_{2}}{T_{1} - T_{2}}

l = \frac{l_{1} T_{2} - l_{2} T_{1}}{T_{2} - T_{1}}

Question 25. The temperature of a wire of length 1 m and area of cross section

1 c m^{2}

is increased from

0^{\circ} C

100^{\circ} C

. If the length of rod is not allowed to increase, then force developed will be

(α = 10^{- 5} / C a n d Y = 10^{11} \frac{N}{m^{2}})

103 N
104 N
105 N
109 N

Discuss Question

Answer: Option B. -> 104 N
:
B
F (force developed)

= Y \propto △ θ

= 10^{11} \times 10^{- 4} \times 10^{- 5} \times 100 = 10^{4} N

Question 26. A 100 N force stretches the length of a hanging wire by 0.5 mm. The force required to stretch a wire, of the same material and length but having four times the diameter, by 0.5 mm is

100 N
400 N
1200 N
1600 N

Discuss Question

Answer: Option D. -> 1600 N
:
D

Y = \frac{F X 4 X 1}{π D^{2} △ l}

In the given problem

F \propto D^{2}

Since Dis increased by a factor of 4 therefore Fis increased by a factor of 16

Question 27. Two identical wires are suspended from the same rigid support but one is of copper and the other is of iron. Young’s modulus of iron is thrice that of copper. The weights to be added on copper and iron wires so that the ends are on the same level must be in the ratio of

1 : 3
2 : 1
3 : 1
4 : 1

Discuss Question

Answer: Option A. -> 1 : 3
:
A

Y \propto F

Therefore

\frac{F_{c u}}{F_{F A}} = \frac{Y_{c u}}{Y_{F A}} = \frac{1}{3}

Question 28. A uniform slender rod of length L, cross sectional area A and Young’s modulus Y is acted upon by the forces shown in the figure. The elongation of the rod is

3FL5AY
2FL5AY
3FL8AY
8FL3AY

Discuss Question

Answer: Option D. -> 8FL3AY
:
D

Net elongation of the rod is

l = \frac{3 F (\frac{2 L}{3})}{A Y} + \frac{2 F (\frac{L}{3})}{A Y}

l = \frac{8 F L}{3 A Y}

Question 29. The Young’s modulus of brass and steel are

10 \times 10^{10} N m^{- 2} a n d 2 \times 10^{11} N m^{- 2}

respectively. A brass wire and a steel wire of the same length are extended by 1 mm under the same force. The radii of the brass and steel wires are

R_{B}

and

R_{s}

respectively. Then

RA=√2RB
RS=RB√2
RS=4RB
RS=RB4

Discuss Question

Answer: Option B. -> RS=RB√2
:
B

Y = \frac{F}{π R^{2}} \times \frac{1}{△ l}

F, l and

△ l

are constants
Therefore

R^{2} \propto \frac{1}{Y}

\frac{R_{S}^{2}}{R_{B}^{2}} = \frac{Y_{B}}{Y_{S}} = \frac{10^{11}}{2 \times 10^{11}} = \frac{1}{2}

o r \frac{R_{S}}{R_{B}} = \frac{1}{\sqrt{2}} o r R_{s} = \frac{R_{B}}{\sqrt{2}}

Question 30. A thick rope of rubber of density

1.5 \times 10^{3} k g m^{- 3}

and Young’s modulus

5 \times 10^{6} N m^{- 2},

8m in length is hung from the ceiling of a room, the increase in its length due to its own weight is

9.6X 10−2 m
19.2 × 10−2 m
9.6× 10−3 m
9.6 m

Discuss Question

Answer: Option A. -> 9.6X 10−2 m
:
A
If A is the area of cross section, d be the density of material and l is lengthof rope
then mass of ropem=Ald
As the weight of the rope acts onthe mid point of the rope