1
MCQ (Single Correct Answer)

JEE Main 2016 (Online) 9th April Morning Slot

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

V(z) = 30 $$-$$ 5x2 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.
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

Explanation

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

$$ \Rightarrow $$   $$\rho $$ = $${10\,{\varepsilon _0}}$$
2
MCQ (Single Correct Answer)

JEE Main 2016 (Online) 10th April Morning Slot

Within a spherical charge distribution of charge density $$\rho $$(r), N equipotential surfaces of potential V0, V0 + $$\Delta $$V, V0 + 2$$\Delta $$V, .......... V0 + N$$\Delta $$V ($$\Delta $$ V > 0), are drawn and have increasing radii r0, r1, r2,..........rN, respectively. If the difference in the radii of the surfaces is constant for all values of V0 and $$\Delta $$V then :
A
$$\rho $$ (r) $$\alpha $$ r
B
$$\rho $$ (r) = constant
C
$$\rho $$ (r) $$\alpha $$ $${1 \over r}$$
D
$$\rho $$ (r) $$\alpha $$ $${1 \over {{r^2}}}$$

Answer

$$\Delta $$

Explanation



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) $$ \propto $$ $${1 \over r}$$
3
MCQ (Single Correct Answer)

JEE Main 2016 (Online) 10th April Morning Slot

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

Explanation

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

CAB = $${{C \times {{32} \over 9}} \over {C + {{32} \over 9}}}$$

Given that,

CAB = 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
4
MCQ (Single Correct Answer)

JEE Main 2017 (Offline)

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

Explanation

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.

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

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

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

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