Joint Entrance Examination

Graduate Aptitude Test in Engineering

Strength of Materials Or Solid Mechanics

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Engineering Mathematics

General Aptitude

1

Within a spherical charge distribution of charge density $$\rho $$(r), N equipotential surfaces of potential V_{0}, V_{0} + $$\Delta $$V, V_{0} + 2$$\Delta $$V, .......... V_{0} + N$$\Delta $$V ($$\Delta $$ V > 0), are drawn and have increasing radii r_{0}, r_{1}, r_{2},..........r_{N}, respectively. If the difference in the radii of the surfaces is constant for all values of V_{0} and $$\Delta $$V then :

A

$$\rho $$ (r) $$\alpha $$ r

B

$$\rho $$ (r) = constant

C

$$\rho $$ (r) $$\alpha $$ $${1 \over r}$$

D

$$\rho $$ (r) $$\alpha $$ $${1 \over {{r^2}}}$$

$$\Delta $$

Here, $$\Delta $$v and $$\Delta $$r are same for any pair of surface.

we know,

Electric field, E = $$-$$ $${{dv} \over {dr}}$$

$$ \therefore $$ E = constant [As dv and dr are constant]

Electric field inside the spherical charge distribution.

E = $${{\rho r} \over {3{\varepsilon _0}}}$$

Now, as E = constant

$$ \therefore $$ $$\rho $$ r = constant

$$ \Rightarrow $$ $$\rho $$ (r) $$ \propto $$ $${1 \over r}$$

2

Figure shows a network of capacitors where the numbers indicates capacitances in micro Farad. The value of capacitance C if the equivalent capacitance between point A and B is to be 1 $$\mu $$F is :

A

$${{31} \over {23}}\,\mu F$$

B

$${{32} \over {23}}\,\mu F$$

C

$${{33} \over {23}}\,\mu F$$

D

$${{34} \over {23}}\,\mu F$$

Equivalent capacitance of 6 $$\mu $$F and 12 $$\mu $$F is = $${{6 \times 12} \over {12 + 6}}$$ = 4 $$\mu $$F

Equivalent capacitance of 4$$\mu $$F and 4$$\mu $$F

is = 4 + 4 = 8 $$\mu $$F

New circuit is $$ \to $$

equivalent capacitance of 1 $$\mu $$F and 8 $$\mu $$F is

= $${{1 \times 8} \over {8 + 1}}$$ = $${8 \over 9}$$ $$\mu $$F

Equivalent capacitance of 8 $$\mu $$F and 4 $$\mu $$F is

= $${{8 \times 4} \over {12}}$$ = $${8 \over 3}$$ $$\mu $$F

New circuit is $$ \to $$

Equation capacitance of $${8 \over 9}$$ $$\mu $$F and $${8 \over 3}$$ $$\mu $$F is

= $${8 \over 9}$$ + $${8 \over 3}$$ = $${32 \over 9}$$ $$\mu $$F

$$ \therefore $$ Equivalent capacitance of AB is

C_{AB} = $${{C \times {{32} \over 9}} \over {C + {{32} \over 9}}}$$

Given that,

C_{AB} = 1 $$\mu $$F

$$ \therefore $$ 1 = $${{C \times {{32} \over 9}} \over {C + {{32} \over 9}}}$$

$$ \Rightarrow $$ C + $${{{32} \over 9}}$$ = $${{32C} \over 9}$$

$$ \Rightarrow $$ $${{23C} \over 9} = {{32} \over 9}$$

$$ \Rightarrow $$ C = $${{32} \over {23}}$$ $$\mu $$F

Equivalent capacitance of 4$$\mu $$F and 4$$\mu $$F

is = 4 + 4 = 8 $$\mu $$F

New circuit is $$ \to $$

equivalent capacitance of 1 $$\mu $$F and 8 $$\mu $$F is

= $${{1 \times 8} \over {8 + 1}}$$ = $${8 \over 9}$$ $$\mu $$F

Equivalent capacitance of 8 $$\mu $$F and 4 $$\mu $$F is

= $${{8 \times 4} \over {12}}$$ = $${8 \over 3}$$ $$\mu $$F

New circuit is $$ \to $$

Equation capacitance of $${8 \over 9}$$ $$\mu $$F and $${8 \over 3}$$ $$\mu $$F is

= $${8 \over 9}$$ + $${8 \over 3}$$ = $${32 \over 9}$$ $$\mu $$F

$$ \therefore $$ Equivalent capacitance of AB is

C

Given that,

C

$$ \therefore $$ 1 = $${{C \times {{32} \over 9}} \over {C + {{32} \over 9}}}$$

$$ \Rightarrow $$ C + $${{{32} \over 9}}$$ = $${{32C} \over 9}$$

$$ \Rightarrow $$ $${{23C} \over 9} = {{32} \over 9}$$

$$ \Rightarrow $$ C = $${{32} \over {23}}$$ $$\mu $$F

3

A capacitance of 2 $$\mu $$F is required in an electrical circuit across a potential difference of 1.0 kV. A large
number of 1 $$\mu $$F capacitors are available which can withstand a potential difference of not more than 300 V.
The minimum number of capacitors required to achieve this is:

A

2

B

16

C

32

D

24

To get a capacitance of 2 μF arrangement of capacitors of capacitance 1μF as shown in figure
8 capacitors of 1μF in parallel with four such branches in series i.e., 32 such capacitors are
required.

$${1 \over {{C_{eq}}}} = {1 \over 8} + {1 \over 8} + {1 \over 8} + {1 \over 8}$$

$$ \Rightarrow $$ $${1 \over {{C_{eq}}}} = {1 \over 2}$$

$$ \Rightarrow $$ $${{C_{eq}} = 2}$$

$${1 \over {{C_{eq}}}} = {1 \over 8} + {1 \over 8} + {1 \over 8} + {1 \over 8}$$

$$ \Rightarrow $$ $${1 \over {{C_{eq}}}} = {1 \over 2}$$

$$ \Rightarrow $$ $${{C_{eq}} = 2}$$

4

An electric dipole has a fixed dipole moment $$\overrightarrow p $$, which makes angle $$\theta$$ with respect to x-axis. When
subjected to an electric field $$\mathop {{E_1}}\limits^ \to = E\widehat i$$ , it experiences a torque $$\overrightarrow {{T_1}} = \tau \widehat k$$ . When subjected to another electric
field $$\mathop {{E_2}}\limits^ \to = \sqrt 3 {E_1}\widehat j$$ it experiences a torque $$\mathop {{T_2}}\limits^ \to = \mathop { - {T_1}}\limits^ \to $$ . The angle $$\theta$$ is:

A

90^{o}

B

45^{o}

C

30^{o}

D

60^{o}

Torque experienced by the dipole in an electric field,

$$T $$ = pE sin$$\theta $$

$$\overrightarrow T = \overrightarrow p \times \overrightarrow E $$

$$\overrightarrow p = p\cos \theta \widehat i + p\sin \theta \widehat j$$

$$\mathop {{E_1}}\limits^ \to = E\widehat i$$

$$\overrightarrow {{T _1}} = \overrightarrow P \times {\overrightarrow E _1}$$

= ($$p\cos \theta \widehat i + p\sin \theta \widehat j$$) $$ \times $$ $$E\left( {\widehat i} \right)$$

= pE sin$$\theta $$$$\left( { - \widehat k} \right)$$

$$\mathop {{E_2}}\limits^ \to = \sqrt 3 {E_1}\widehat j$$

$$\overrightarrow {{T _2}} = $$($$p\cos \theta \widehat i + p\sin \theta \widehat j$$) $$ \times $$ $$\sqrt 3 {E_1}\widehat j$$

= $$\sqrt 3 pE\cos \theta \left( {\widehat k} \right)$$

Now given, $$\overrightarrow {{T _2}}$$ = $$-\overrightarrow {{T _1}}$$

$$ \Rightarrow $$ $$\sqrt 3 pE\cos \theta \left( {\widehat k} \right)$$ = -pE sin$$\theta $$$$\left( { - \widehat k} \right)$$

$$ \Rightarrow $$ $$\tan \theta = \sqrt 3 $$

$$ \Rightarrow $$ $$\theta $$ = 60

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