Joint Entrance Examination

Graduate Aptitude Test in Engineering

1

Numerical

A steel rod with y = 2.0 $$\times$$ 10^{11} Nm^{$$-$$2} and $$\alpha$$ = 10^{$$-$$5} $$^\circ$$C^{$$-$$1} of length 4 m and area of cross-section 10 cm^{2} is heated from 0$$^\circ$$C to 400$$^\circ$$C without being allowed to extend. The tension produced in the rod is x $$\times$$ 10^{5} N where the value of x is ____________.

Your Input ________

Correct Answer is **8**

Given, the Young's modulus of the steel rod, Y = 2 $$\times$$ 10^{11} Pa

Thermal coefficient of the steel rod, $$\alpha$$ = 10^{$$-$$5}$$^\circ$$C

The length of the steel rod, l = 4 m

The area of the cross-section, A = 10 cm^{2}

The temperature difference, $$\Delta$$T = 400$$^\circ$$C

As we know that,

Thermal strain = $$\alpha$$ $$\Delta$$T

Using the Hooke's law

Young's modulus (Y) = $${{Thermal\,stress} \over {Thermal\,strain}} = {{F/A} \over {\alpha \,\Delta \,T}}$$

Thermal stress, $$F = YA\,\alpha \,\Delta \,T$$

Substitute the values in the above equation, we get

$$F = 2 \times {10^{11}} \times 10 \times {10^{ - 4}} \times {10^{ - 5}} \times (400)$$

$$ = 8 \times {10^5}N$$

Comparing with, $$F = x \times {10^5}N$$

The value of the x = 8.

Thermal coefficient of the steel rod, $$\alpha$$ = 10

The length of the steel rod, l = 4 m

The area of the cross-section, A = 10 cm

The temperature difference, $$\Delta$$T = 400$$^\circ$$C

As we know that,

Thermal strain = $$\alpha$$ $$\Delta$$T

Using the Hooke's law

Young's modulus (Y) = $${{Thermal\,stress} \over {Thermal\,strain}} = {{F/A} \over {\alpha \,\Delta \,T}}$$

Thermal stress, $$F = YA\,\alpha \,\Delta \,T$$

Substitute the values in the above equation, we get

$$F = 2 \times {10^{11}} \times 10 \times {10^{ - 4}} \times {10^{ - 5}} \times (400)$$

$$ = 8 \times {10^5}N$$

Comparing with, $$F = x \times {10^5}N$$

The value of the x = 8.

2

Numerical

When a rubber ball is taken to a depth of __________m in deep sea, its volume decreases by 0.5%. (The bulk modulus of rubber = 9.8 $$\times$$ 10^{8} Nm^{$$-$$2}, Density of sea water = 10^{3} kgm^{$$-$$3}, g = 9.8 m/s^{2})

Your Input ________

Correct Answer is **500**

$$B = - {{\Delta P} \over {\left( {{{\Delta V} \over V}} \right)}} = - {{\rho gh} \over {\left( {{{\Delta V} \over V}} \right)}}$$

$$ - {{B{{\Delta V} \over V}} \over {\rho g}} = h$$

$${{9.8 \times {{10}^8} \times 0.5} \over {100 \times {{10}^3} \times 9.8}} = h$$

h = 500

$$ - {{B{{\Delta V} \over V}} \over {\rho g}} = h$$

$${{9.8 \times {{10}^8} \times 0.5} \over {100 \times {{10}^3} \times 9.8}} = h$$

h = 500

3

Numerical

Wires W_{1} and W_{2} are made of same material having the breaking stress of 1.25 $$\times$$ 10^{9} N/m^{2}. W_{1} and W_{2} have cross-sectional area of 8 $$\times$$ 10^{$$-$$7} m^{2} and 4 $$\times$$ 10^{$$-$$7} m^{2}, respectively. Masses of 20 kg and 10 kg hang from them as shown in the figure. The maximum mass that can be placed in the pan without breaking the wires is ____________ kg. (Use g = 10 m/s^{2})

Your Input ________

Correct Answer is **40**

B.S_{1} = $${{{T_{1\max }}} \over {8 \times {{10}^{ - 7}}}}$$ $$\Rightarrow$$ T_{1 max} = 8 $$\times$$ 1.25 $$\times$$ 100 = 1000 N

B.S_{2} = $${{{T_{2\max }}} \over {4 \times {{10}^{ - 7}}}}$$ $$\Rightarrow$$ T_{2 max} = 4 $$\times$$ 1.25 $$\times$$ 100 = 500 N

m = $${{500 - 100} \over {10}} = 40$$ kg

B.S

m = $${{500 - 100} \over {10}} = 40$$ kg

4

Numerical

A soap bubble of radius 3 cm is formed inside the another soap bubble of radius 6 cm. The radius of an equivalent soap bubble which has the same excess pressure as inside the smaller bubble with respect to the atmospheric pressure is ................ cm.

Your Input ________

Correct Answer is **2**

Image

Excess pressure inside the smaller soap bubble

$$\Delta P = {{4S} \over {{r_1}}} + {{4S} \over {{r_2}}}$$ .... (i)

The excess pressure inside equivalent soap bubble

$$\Delta P = {{4S} \over {{R_{eq}}}}$$ ....... (ii)

From (i) & (ii)

$${{4S} \over {{R_{eq}}}} = {{4S} \over {{r_1}}} + {{4S} \over {{r_2}}}$$

$${1 \over {{R_{eq}}}} = {1 \over {{r_1}}} + {1 \over {{r_2}}}$$

$$ = {1 \over 6} + {1 \over 3}$$

R_{eq} = 2 cm

Excess pressure inside the smaller soap bubble

$$\Delta P = {{4S} \over {{r_1}}} + {{4S} \over {{r_2}}}$$ .... (i)

The excess pressure inside equivalent soap bubble

$$\Delta P = {{4S} \over {{R_{eq}}}}$$ ....... (ii)

From (i) & (ii)

$${{4S} \over {{R_{eq}}}} = {{4S} \over {{r_1}}} + {{4S} \over {{r_2}}}$$

$${1 \over {{R_{eq}}}} = {1 \over {{r_1}}} + {1 \over {{r_2}}}$$

$$ = {1 \over 6} + {1 \over 3}$$

R

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