Free Objective Test 01 Practice Test - 11th and 12th

Question 1

The ratio of diameters of two wires of same materials is n : 1.The length of each wire is 4 m. On applying the same load, the increase in length of thin wire will be $(n > 1)$

n^{2}

times

B. n times

C. 2n times

D. (2n + 1) times

SOLUTION

Solution : A

$Y = \frac{\frac{F}{A}}{\frac{△ l}{l}} = \frac{F l}{A △ l}$
or $Y = \frac{F l X 4}{π D^{2} X △ l} o r △ l \propto \frac{1}{D^{2}} o r \frac{△ L_{2}}{△ L_{1}} = \frac{D_{1}^{2}}{D_{1}^{2}} = \frac{n^{2}}{1}$

Question 2

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

9.6 X 10^{- 2}

19.2 \times 10^{- 2}

9.6 \times 10^{- 3}

D. 9.6 m

SOLUTION

Solution : A

If A is the area of cross section, d be the density of material and l is length of rope
then mass of rope m=Ald
As the weight of the rope acts on the mid point of the rope
$Y = \frac{m g}{A} \times \frac{l}{△ l}$
$△ l = \frac{m g l}{2 A Y} = \frac{A l p g l}{2 A Y} = \frac{g p l^{2}}{2 Y}$
or $△ l = \frac{9.8 \times 1.5 \times 10^{3} \times 8^{2}}{2 \times 5 \times 10^{6}} = 9.6 \times 10^{- 2} m$

Question 3

The diameter of a brass wire is 0.6 mm and Y is $9 \times 10^{6} N m^{- 2} .$ The force which will increase its length by 0.2% is about

A. 100 N

B. 51 N

C. 25 N

D. none of these

SOLUTION

Solution : B

$F = \frac{Y A △ l}{l}$
$= 9 \times 10^{10} \times \frac{22}{7} \times \frac{(0.6 \times 10^{- 3})^{2}}{4} \times \frac{0.2}{100} N$ =51 N approx

Question 4

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

A. 1 : 3

B. 2 : 1

C. 3 : 1

D. 4 : 1

SOLUTION

Solution : A

$Y \propto F$
Therefore $\frac{F_{c u}}{F_{F A}} = \frac{Y_{c u}}{Y_{F A}} = \frac{1}{3}$

Question 5

The temperature of a wire of length 1 m and area of cross section
$1 c m^{2}$ is increased from $0^{\circ} C$ to $100^{\circ} C$ . If the length of rod is not allowed to increase, then force developed will be $(α = 10^{- 5} / C a n d Y = 10^{11} \frac{N}{m^{2}})$

10^{3}

10^{4}

10^{5}

10^{9}

SOLUTION

Solution : B

F (force developed) $= Y \propto △ θ$
$= 10^{11} \times 10^{- 4} \times 10^{- 5} \times 100 = 10^{4} N$

Question 6

Two wires of the same material and length are stretched by the same force. Their masses are in the ratio 3:2. Their elongations are in the ratio

A. 3 : 2

B. 9 : 4

C. 2 : 3

D. 4 : 9

SOLUTION

Solution : C

$Y = \frac{F l}{A △ l} o r △ l \propto \frac{1}{A}$
Again, $m = A l p, m \propto A$
$△ l \propto \frac{1}{m}$
Therefore $\frac{△ l_{1}}{△ l_{2}} = \frac{m_{2}}{m_{1}} = \frac{2}{3}$

Question 7

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

A. 100 N

B. 400 N

C. 1200 N

D. 1600 N

SOLUTION

Solution : D

$Y = \frac{F X 4 X 1}{π D^{2} △ l}$

In the given problem $F \propto D^{2}$

Since D is increased by a factor of 4 therefore F is increased by a factor of 16

Question 8

Two wires of the same material have lengths in the ratio 1 : 2 and their radii are in the ratio 1 : $\sqrt{2}$ . If they are stretched by applying equal forces, the increase in their lengths will be in the ratio

\sqrt{2} : 2

2 : \sqrt{2}

C. 1 : 1

D. 1 : 2

SOLUTION

Solution : C

$Y = \frac{F l}{π r^{2} △ l} o r △ l = \frac{F}{π r^{2} y}$
$△ l \propto \frac{1}{r^{2}}$
$△ l^{'} \propto \frac{21}{(\sqrt{2} r)^{2}} o r △ l^{'} \propto \frac{1}{r^{2}}$
Therefore $\frac{△ l}{△ l^{'}}$ =1

Question 9

An aluminium rod, Young’s modulus $7 \times 10^{9} N m^{- 2},$ has a breaking strain of 0.2%. The minimum cross sectional area of the rod in $m^{2}$ in order to support a load of
$10^{4}$ N is

1 \times 10^{- 2}

1.4 \times 10^{- 3}

1 \times 10^{- 3}

7.1 \times 10^{- 4}

SOLUTION

Solution : D

$Y = \frac{\frac{F}{A}}{B r e a k i n g s t r a i n}$

or $a = \frac{F}{Y \times B r e a k i n g s t r a i n} = \frac{10^{4} \times 100}{7 \times 10 \times 0.2}$
$= 0.71 \times 10^{- 3} = 7.1 \times 10^{- 4}$

Question 10

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

A. 8

B. 4

C. 2

D. 1

SOLUTION

Solution : A

$Y = \frac{F l}{A △ l} o r △ l \propto \frac{F}{r^{2}}$

or $\frac{△ l_{2}}{△ l_{1}} = \frac{F_{2}}{F_{1}} \times \frac{r_{1}^{2}}{r_{2}^{2}}$

or $\frac{△ l_{2}}{△ l_{1}} = 2 \times 2 \times 2 = 8$

or $△ l_{2} = 8 △ l_{1} = 8 \times 1 = 8$ mm

Question 11

A wire is stretched 1 mm by a force of 1 KN. How far would a wire of the same material and length but of four times that diameter be stretched by the same force?

\frac{1}{2}

\frac{1}{2}

\frac{1}{8}

\frac{1}{16}

SOLUTION

Solution : D

$Y = \frac{F l}{A △ l}$
Y,l and F are constants
Therefore $△ l \propto \frac{1}{D^{2}}$

$\frac{△ l_{2}}{△ l_{1}} = \frac{D_{1}^{2}}{D_{2}^{2}} = \frac{1}{16}$

Therefore $△ l_{2} = \frac{1}{16}$ mm

Question 12

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

R_{A} = \sqrt{2} R_{B}

R_{S} = \frac{R_{B}}{\sqrt{2}}

R_{S} = 4 R_{B}

R_{S} = \frac{R_{B}}{4}

SOLUTION

Solution : 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 13

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

\frac{T_{2}}{T_{1}} (l_{2} + l_{2})

T_{1} l_{1} + T_{2} l_{2}

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

\frac{l_{1} T_{2} + l_{2} T_{1}}{T_{2} + T_{1}}

SOLUTION

Solution : 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, $\frac{l_{1} - l}{l_{2} - l} = \frac{T_{1}}{T_{2}}$
or $l_{1} T_{2} - l T_{2} = l_{2} T_{1} - l T_{1}$
or $l (T_{1} - T_{2}) = l_{2} T_{1} - l_{1} T_{2}$

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

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

Question 14

Two identical wires of rubber and iron are stretched by the same weight, then the number of atoms in the iron wire will be

A. equal to that of rubber

B. less than that of the rubber

C. more than that of rubber

D. none of the above

SOLUTION

Solution : C

Iron is more elastic than rubber, hence the number of atoms per unit volume will be more in case of iron.

Question 15

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

\frac{3 F L}{5 A Y}

\frac{2 F L}{5 A Y}

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

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

SOLUTION

Solution : 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 16

Two wires of the same material (Young’s modulus Y) and same length L but radii R and 2R respectively are joined end to end and a weight w is suspended from the combination as shown in the figure. The elastic potential energy in the system is

\frac{3 w^{2} L}{4 π R^{2} Y}

\frac{3 w^{2} L}{8 π R^{2} Y}

\frac{5 w^{2} L}{8 π R^{2} Y}

\frac{w^{2} L}{π R^{2} Y}

SOLUTION

Solution : C

$K_{1} = \frac{y π (2 R)^{2}}{L}$
$K_{2} = \frac{y π (R)^{2}}{L}$
Equivalent $\frac{1}{K_{1}} + \frac{1}{K_{2}} = \frac{1}{4 y π R^{2}} + \frac{L}{y π R^{2}}$
Since, $K_{1} x_{1} = K_{2} x_{2} = w$
Elastic potential energy of the system
U= $\frac{1}{2} k_{1} x_{1} + \frac{1}{2} k_{2} x_{2}^{2}$
$= \frac{1}{2} k_{1} (\frac{W}{k_{1}})^{2} + \frac{1}{2} k_{2} (\frac{W}{k_{2}})^{2}$
$= \frac{1}{2} W^{2} + (\frac{1}{k_{1}} + \frac{1}{k_{2}})$
$= \frac{1}{2} W^{2} + (\frac{5 L}{4 y π R^{2}})$

U= $(\frac{5 w^{2} L}{8 π y R^{2}})$

Question 17

A cube is compressed at $0^{\circ} C$ equally from all sides by an external pressure p. By what amount should the temperature be raised to bring it back to the size it had before the external pressure was applied? (Given K is bulk modulus of elasticity of the material of the cube and $α$ is the coefficient of linear expansion)

\frac{p}{K α}

\frac{p}{3 K α}

\frac{3 π α}{p}

\frac{K}{3 p}

SOLUTION

Solution : B

$k = \frac{p V}{△ v} = \frac{p V}{y △ T} = \frac{p}{3 α T}$
or $T = \frac{p}{3 k α}$

Question 18

Two rods A and B of the same material and length have their radii $r_{1}$ and $r_{2}$ respectively. When they are rigidly fixed at one end and twisted by the same couple applied at the other end, the ratio of the angle of twist at the end of A and the angle of twist at the end of B is

\frac{r_{2}^{4}}{r_{1}^{4}}

\frac{r_{1}^{4}}{r_{2}^{4}}

\frac{r_{2}^{2}}{r_{1}^{2}}

\frac{r_{1}^{2}}{r_{2}^{2}}

SOLUTION

Solution : A

$τ = \frac{π n r^{4}}{21} θ$
In the given problem, $r^{4} θ =$ constant
Therefore $\frac{θ_{A}}{θ_{B}} = \frac{r_{2}^{4}}{r_{1}^{4}}$

Question 19

What amount of work is done in increasing the length of a wire through unity?

\frac{Y L}{2 A}

\frac{Y L^{2}}{2 A}

\frac{Y A}{2 L}

\frac{Y L}{A}

SOLUTION

Solution : C

Work done $= \frac{1}{2} F \times$ extension

$= \frac{1}{2} \times \frac{Y A}{L} \times 1 y = \frac{F \times L}{A \times 1}$

$= \frac{Y A}{2 L} h e n c e F = \frac{Y A}{2 L}$

Question 20

Two wires of same material and radius have their lengths in ratio 1 : 2. If these wires are stretched by the same force, the strain produced in the two wires will be in the ratio

A. 2 : 1

B. 1 : 1

C. 1 : 2

D. 1 : 4

SOLUTION

Solution : C

l= $\frac{F L}{A Y}$

l= $\propto \frac{1}{r^{2}}$

$\frac{l_{1}}{l_{2}} = \frac{L_{1}}{L_{2}} \times \frac{r_{2}^{2}}{r_{1}^{2}}$

$o r \frac{l_{1}}{l_{2}} = \frac{1}{2}$

Free Objective Test 01 Practice Test - 11th and 12th

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Question 11

SOLUTION

Question 12

SOLUTION

Question 13

SOLUTION

Question 14

SOLUTION

Question 15

SOLUTION

Question 16

SOLUTION

Question 17

SOLUTION

Question 18

SOLUTION

Question 19

SOLUTION

Question 20

SOLUTION

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