Mechanical Properties of Solids Test

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Mechanical Properties of Solids Test - 38

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Weekly Quiz Competition

Question 1
1 / -0

The maximum load that a wire can sustain is $$W$$. If the wire is cut to half its value, the maximum load itcan sustain is

A
$$W$$

B
$$\dfrac{w}{2}$$

C
$$\dfrac{w}{4}$$

D
$$2W$$

Solution

stress $$= \dfrac{F}{A}$$
$$F = stress \times A$$
still the value of maximum force or load remains the same as it dod'nt depend on length of wire.
Question 2
1 / -0

A uniform steel rod of cross-sectional area $$A$$ and length $$L$$ is suspended so that it hangs vertically. The stress at the middle point of the rod is :

A
$$\dfrac{1}{2}\rho gL$$

B
$$\dfrac{1}{4}\rho gL$$

C
$$\rho gL$$

D
None of these

Solution

Force acting at the mid point of the rod is
$$T=\dfrac {M}{2}(g)=\dfrac {\rho SL}{2}(g)$$
$$\therefore \sigma =\dfrac {T}{S}=\dfrac {1}{2}\rho gl$$
Option A.
Question 3
1 / -0

Identify the case when an elastic metal rod does not undergo elongation:

A
it is pulled with a constant acceleration on a smooth horizontal surface

B
it is pulled with constant velocity on a rough horizontal surface

C
it is allowed to fall freely

D
all of the above

Solution

To have no net elongation in the rod , no force should act on it. In case of free fall, every point have some acceleration. So, there will not be any elongation.
Question 4
1 / -0

The bulk modulus of water is $$2.0 \times 10^{9} N/m^{2}$$ The pressure required to increase the density of water by $$0.1\%$$ is :

A
$$2.0\times 10^{3} N/m^{2}$$

B
$$2.0\times 10^{6} N/m^{2}$$

C
$$2.0\times 10^{5} N/m^{2}$$

D
$$2.0\times 10^{7} N/m^{2}$$

Solution

$$\rho^{'} = \rho(1+\dfrac{dp}{B})$$
change in density $$\Delta \rho =\rho^{'} -\rho = \dfrac{\rho dp}{B}$$
$$\dfrac{\Delta \rho}{\rho} = \dfrac{dp}{B}$$
$$\dfrac{\Delta \rho}{\rho} \times 100 = 0.1$$
$$\dfrac{\Delta \rho}{\rho} = 0.001$$
$$\Delta P = B\dfrac{\Delta \rho}{\rho} = 2 \times 10^9 \times 0.001 = 2\times 10^6 N/m^2$$
Question 5
1 / -0

Vessel of $$1 \times 10^{-3} m^{3}$$volume contains an oi. If a pressure of $$1.2 \times 10^{5} N/m^{2}$$is applied on it, thenvolume decreases by $$0.3\times 10^{-3}m^{3}$$ . The bulk modulus of oil is

A
$$6\times 10^{10}N/m^{2}$$

B
$$4\times 10^{5}N/m^{2}$$

C
$$2\times 10^{7}N/m^{2}$$

D
$$1\times 10^{6}N/m^{2}$$

Solution

Bulk modulus of the oil is given by
$$B=-\dfrac {\Delta P}{\Delta V/V}=-\dfrac {(1.2\times 10^5)(1\times 10^{-3})}{-0.3\times 10^{-3}}=4\times 10^5N/m^2$$
Option B.
Question 6
1 / -0

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 :

A
$$2$$

B
$$\sqrt{2}:2$$

C
$$1:1$$

D
$$1:2$$

Solution

$$\displaystyle \Delta\ L =\frac{Fl}{YA}; \ \frac{\Delta l}{\Delta l_{2}}=\frac{l_{1}}{l_{2}}\times \left ( \frac{r_{2}}{r_{1}} \right )^{2}=\frac{1}{2}\times \left ( \frac{\sqrt{2}}{1} \right )^{2}=1:1$$
Question 7
1 / -0

The area of cross-section of a wire of length $$1.1$$ meter is $$1\:mm^{2}$$. It is loaded with 1 kg. If Young's modulus of copper is $$1.1\times 10^{11}N/m^{2}$$, then the increase in length will be (If $$g=10\:m/s^{2}$$) :

A
$$0.01\ mm$$

B
$$0.075 \ mm$$

C
$$0.1\ mm$$

D
$$0.15\ mm$$

Solution

$$\displaystyle y=\dfrac{F/A}{\dfrac{\Delta l}{l}}$$
$$\displaystyle \Delta l=\frac{Fl}{Ay}$$
$$\displaystyle \Delta l=\frac{Mgl}{Ay}$$
$$\displaystyle \Delta l=\frac{1\times 10\times 1.1}{1\times 10^{-6}\times 1.1\times 10^{11}}=1\times 10^{-4}=0.1\:mm$$
Question 8
1 / -0

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 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

Solution

$$\displaystyle Y=\frac{F/A}{\frac{\Delta l}{l}}$$ $$\displaystyle F=\frac{YA\Delta l}{l}$$
$$\displaystyle F=Y.\pi r^{2}\frac{\Delta l}{l}$$
$$\displaystyle F=0.9\times 10^{11}\times 3.14\times \left ( \frac{0.6}{2}\times 10^{-3} \right )^{2}\times \frac{0.2}{100}$$
$$\displaystyle F=0.9\times 10^{11}\times 3.14\times \times 0.09\times 10^{-6}\times 2\times 10^{-3}$$
$$\displaystyle F=51\:N$$
Question 9
1 / -0

A force of $$10^{3}$$ N stretches the length of a hanging wire by $$1$$ millimeter. The force required to stretch a wire of same material and length but having four times the diameter by $$1$$ millimeter is :

A
$$4\times 10^{3}$$ N

B
$$16\times 10^{3}$$ N

C
$$\displaystyle \frac{1}{4}\times 10^{3}$$ N

D
$$\displaystyle \frac{1}{16}\times 10^{3}$$ N

Solution

$$\displaystyle Y=\frac{F/A}{\Delta l/l}$$ ; $$Y$$ & $$\displaystyle \frac{\Delta l}{l}=constant$$
$$\displaystyle F\propto A\propto r^{2}$$ $$\displaystyle \frac{F_{2}}{F_{1}}=\left ( \frac{r_{2}}{r_{1}} \right )^{2}=\left ( \frac{4r}{r} \right )^{2}=16$$
$$F_{2}=16\times 10^{3}\:N$$
Question 10
1 / -0

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

A
$$\displaystyle \dfrac{1}{l^{2}}$$

B
$$\displaystyle \dfrac{1}{l}$$

C
$$l^{2}$$

D
$$l$$

Solution

let the volume be $$V$$ and area $$A$$

$$Y = \dfrac{Fl}{A \Delta l}$$

$$\Delta l = \dfrac{Fl}{YA}$$

$$V = A \times l$$

$$A =\dfrac{V}{l}$$

$$\Delta l = \dfrac{Fl^2}{YV}$$

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Mechanical Properties of Solids Test - 38

Result

Mechanical Properties of Solids Test - 38

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SHARING IS CARING

Question 1

$$W$$

$$\dfrac{w}{2}$$

$$\dfrac{w}{4}$$

$$2W$$

Solution

Question 2

$$\dfrac{1}{2}\rho gL$$

$$\dfrac{1}{4}\rho gL$$

$$\rho gL$$

None of these

Solution

Question 3

it is pulled with a constant acceleration on a smooth horizontal surface

it is pulled with constant velocity on a rough horizontal surface

it is allowed to fall freely

all of the above

Solution

Question 4

$$2.0\times 10^{3} N/m^{2}$$

$$2.0\times 10^{6} N/m^{2}$$

$$2.0\times 10^{5} N/m^{2}$$

$$2.0\times 10^{7} N/m^{2}$$

Solution

Question 5

$$6\times 10^{10}N/m^{2}$$

$$4\times 10^{5}N/m^{2}$$

$$2\times 10^{7}N/m^{2}$$

$$1\times 10^{6}N/m^{2}$$

Solution

Question 6

$$2$$

$$\sqrt{2}:2$$

$$1:1$$

$$1:2$$

Solution

Question 7

$$0.01\ mm$$

$$0.075 \ mm$$

$$0.1\ mm$$

$$0.15\ mm$$

Solution

Question 8

Nearly 17 N

Nearly 34 N

Nearly 51 N

Nearly 68 N

Solution

Question 9

$$4\times 10^{3}$$ N

$$16\times 10^{3}$$ N

$$\displaystyle \frac{1}{4}\times 10^{3}$$ N

$$\displaystyle \frac{1}{16}\times 10^{3}$$ N

Solution

Question 10

$$\displaystyle \dfrac{1}{l^{2}}$$

$$\displaystyle \dfrac{1}{l}$$

$$l^{2}$$

$$l$$

Solution

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