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 2 of 4 pages

Question 11. When a certain weight is suspended from a long uniform wire, its length increases by one cm. If the same weight is suspended from another wire of the same material and length but having a diameter half of the first one then the increase in length will be -

0.5 cm
2 cm
4 cm
8 cm

Discuss Question

Answer: Option C. -> 4 cm
:
C

c) : I = \frac{F L}{A Y} \Rightarrow l \propto \frac{1}{r^{2}}

(F,L and Y are constant)

\frac{l_{2}}{l_{1}} = (\frac{r_{1}}{r_{2}})^{2} = (2)^{2} = 4

\Rightarrow l_{2} = 4 l_{1}

=4cm

Question 12. A wire extends by 1 mm when a force is applied. Double the force is applied to another wire of same material and length but half the radius of cross-section. The elongation of the wire in mm will be

Discuss Question

Answer: Option A. -> 8
:
A

a) I = \frac{F L}{π r^{2} r} l \propto \frac{F}{r^{2}}

(Y and L are constant)

\frac{l_{2}}{l_{1}} = \frac{F_{2}}{F_{1}} \times (\frac{r_{1}}{r_{2}})^{2} = 2 \times (2)^{2} = 8

l_{2} = 8 l_{1} = 8 \times 1 = 8

Question 13. The relationship between Young's modulus Y, Bulk modulus K and modulus of rigidity

η

y=9η kη+3K
y=9YKY+3K
y=9η k3+K
y=3η k9η+K

Discuss Question

Answer: Option A. -> y=9η kη+3K
:
A
Solution: a):

Y = 3 K (1 - 2 σ)

and

y = 2 η (1 + σ)

Eliminating

σ

We get

y = \frac{9 η k}{η + 3 K}

Question 14. A pan with set of weights is attached with a light spring. When disturbed, the mass-spring system oscillates with a time period of 0.6s. When some additional weights are added then time period is 0.7s. The extension caused by the additional weights is approximately

1.38 cm
3.5 cm
1.75 cm
2.45 cm

Discuss Question

Answer: Option B. -> 3.5 cm
:
B
b)2

π \sqrt{\frac{m}{k}}

=0.6....(i) and

2 π \sqrt{\frac{m + m^{'}}{k}} = 0.7 . . . (i i)

Dividing (ii) by (i) we get

(\frac{7}{6})^{2} = \frac{m + m^{'}}{m} = \frac{49}{36}

\frac{m + m^{'}}{m} - 1 = \frac{49}{36} - 1 \Rightarrow \frac{m^{'}}{m} = \frac{13}{36} m^{'} \frac{13 m}{36}

Also

\frac{k}{m} = \frac{4 π^{2}}{(0.6)^{2}}

Desired extension =

\frac{m^{'} g}{k}

\frac{13}{36} \times \frac{m g}{k}

\frac{13}{36} \times 10 \times \frac{0.36}{4 π^{2}} \approx 3.5 c m

Question 15. A student performs an experiment to determine the Young’s modulus of a wire, exactly 2 m long, by Searle’s method. In a particular reading, the students measures the extension in the length of the wire to be 0.8 mm with an uncertainty of +/- 0.05 mm at a load of exactly 1kg. The student also measures the diameter of the wire to be 0.4 mm with an uncertainty of +/-0.01 mm.The Young’s modulus obtained from the reading is

(g = 9.8 m s^{- 2})

(2.0+/−0.3)× 1011 Nm−2
(2.0+/−0.2) × 1011 Nm−2
(2.0+/−0.1)× 1011 Nm−2
(2.0+/−0.05)× 1011 Nm−2

Discuss Question

Answer: Option B. -> (2.0+/−0.2) × 1011 Nm−2
:
B

Y = \frac{4 F l}{π D^{2} △ l}

= \frac{4 \times (1 X 9.8) X 2}{(\frac{22}{7}) (0.4 \times 10^{-} 3)^{2} X (0.8 X 10^{- 3})}

= 2 \times 10^{11} N m^{- 2}

Fractional error in y is

\frac{△ y}{y} = + / - (\frac{2 △ D}{D} + \frac{△ l}{l})

o r △ y = + / - (\frac{2 X 0.01}{0.4} + \frac{0.05}{0.8}) \times 2 \times 10^{11}

= + / - 0.2 \times 10^{11} N m^{- 2}

Therefore

y = (2 \times 10^{11} + / - 0.2 \times 10^{11}) N m^{- 2}

Question 16. A steel ring of radius r and cross-section area 'A' is fitted on to a wooden disc of radius R(R

>

r). If Young's modulus be E, then the force with which the steel ring is expanded is

AERr
AE(R−rr)
EA(R−rA)
ErAR

Discuss Question

Answer: Option C. -> EA(R−rA)
:
C
c): Initial length (circumference) of the ring

= 2 π r

Final length (circumference) of the ring

= 2 π R

Change in length=

2 π R - 2 π r .

strain=

\frac{c h a n g e i n l e n g t h}{o r i g i n a l l e n g t h} = \frac{2 π (R - r)}{2 π r} = \frac{R - r}{r}

Now young's modulus

E = \frac{F / A}{t / L} = \frac{F / A}{(R - r) / r}

F = A E (\frac{R - r}{r})

Question 17. According to Hook's law of elasticity, if stress is increased, the ratio of stress to strain

Increases
Decreases
Becomes zero
Remains constant

Discuss Question

Answer: Option D. -> Remains constant
:
D
(d)Y =

\frac{S t r e s s}{S t r a i n}

=Constant
It depends only on nature of material

Question 18. A wire is stretched under a force. If the wire suddenly snaps, the temperature of the wire

remains the same
decreases
increases
first decreases then increases

Discuss Question

Answer: Option C. -> increases
:
C
The work done on the wire to produce a strain in it will be stored as energy which is converted into heat, when wire snaps suddenly
Due to thisthe temperature increases

Question 19. The pressure of a medium is changed from