Joint Entrance Examination

Graduate Aptitude Test in Engineering

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Engineering Mathematics

General Aptitude

1

Three capacitors each of 4 $$\mu $$F are to be connected in such a way that the effective capacitance is 6 $$\mu $$F. This can be done by connecting them :

A

all in series

B

two in series and one in parallel

C

all in parallel

D

two in parallel and one in series

(a) $${1 \over {{C_{eq}}}} = {1 \over 4} + {1 \over 4} + {1 \over 4} = {3 \over 4}$$

$$ \Rightarrow $$ $${C_{eq}} = {4 \over 3}\,\mu F$$

(b) $${C_{eq}} = {{4 \times 4} \over {4 + 4}} + 4 = 6\mu F$$

(c) $${C_{eq}} = 4 + 4 + 4 = 12\,\mu F$$

(d) $${C_{eq}} = {{\left( {4 + 4} \right) \times 4} \over {\left( {4 + 4} \right) + 4}} = {8 \over 3}\mu F$$

$$ \Rightarrow $$ $${C_{eq}} = {4 \over 3}\,\mu F$$

(b) $${C_{eq}} = {{4 \times 4} \over {4 + 4}} + 4 = 6\mu F$$

(c) $${C_{eq}} = 4 + 4 + 4 = 12\,\mu F$$

(d) $${C_{eq}} = {{\left( {4 + 4} \right) \times 4} \over {\left( {4 + 4} \right) + 4}} = {8 \over 3}\mu F$$

2

The potential (in volts) of a charge distribution is given by.

V(z) = 30 $$-$$ 5x^{2} for $$\left| z \right|$$ $$ \le $$ 1 m.

V(z) = 35 $$-$$ 10 $$\left| z \right|$$ for $$\left| z \right|$$ $$ \ge $$1 m.

V(z) does not depend on x and y. If this potential is generated by a constant charge per unit volume $${\rho _0}$$ (in units of $${\varepsilon _0}$$) which is spread over a certain region, then choose the correct statement.

V(z) = 30 $$-$$ 5x

V(z) = 35 $$-$$ 10 $$\left| z \right|$$ for $$\left| z \right|$$ $$ \ge $$1 m.

V(z) does not depend on x and y. If this potential is generated by a constant charge per unit volume $${\rho _0}$$ (in units of $${\varepsilon _0}$$) which is spread over a certain region, then choose the correct statement.

A

$${\rho _0}$$ = 10 $${\varepsilon _0}$$ for $$\left| z \right|$$ $$ \le $$ 1 m and $${\rho _0} = 0$$ elsewhere

B

$${\rho _0}$$ = 20 $${\varepsilon _0}$$ in the entire region

C

$${\rho _0}$$ = 40 $${\varepsilon _0}$$ in the entire region

D

$${\rho _0}$$ = 20 $${\varepsilon _0}$$ for $$\left| z \right|$$ $$ \le $$ 1 m and $${\rho _0} = 0$$ elsewhere

We know,

E(z) = $$-$$ $${{dv} \over {dz}}$$

$$ \therefore $$ E(z) = $$-$$ 10 z for $$\left| z \right| \le 1$$ m

and E(z) = 10 for $$\left| z \right| \ge 1$$ m

$$ \therefore $$ The source is an infinity large non conducting thick of thickness z = 2 m.

$$ \therefore $$ E = $${\sigma \over {2{\varepsilon _0}}}$$ = $${{\rho \left( 2 \right)} \over {2{\varepsilon _0}}}$$ = $${\rho \over {{\varepsilon _0}}}$$

$$ \therefore $$ $${\rho \over {{\varepsilon _0}}}$$ = 10

$$ \Rightarrow $$ $$\rho $$ = $${10\,{\varepsilon _0}}$$

E(z) = $$-$$ $${{dv} \over {dz}}$$

$$ \therefore $$ E(z) = $$-$$ 10 z for $$\left| z \right| \le 1$$ m

and E(z) = 10 for $$\left| z \right| \ge 1$$ m

$$ \therefore $$ The source is an infinity large non conducting thick of thickness z = 2 m.

$$ \therefore $$ E = $${\sigma \over {2{\varepsilon _0}}}$$ = $${{\rho \left( 2 \right)} \over {2{\varepsilon _0}}}$$ = $${\rho \over {{\varepsilon _0}}}$$

$$ \therefore $$ $${\rho \over {{\varepsilon _0}}}$$ = 10

$$ \Rightarrow $$ $$\rho $$ = $${10\,{\varepsilon _0}}$$

3

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) $$ \propto $$ $${1 \over r}$$

4

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

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