NEET PHYSICS question bank, MCQ CBSE Notes, Lectures

1 Electrostatics

A 10 pF parallel plate capacitor is charged with a 4.0 V battery. While the capacitor is still connected to the battery, a dielectric slab (κ = 5.0) is inserted between the plates to completely fill the gap. How much electric potential energy is stored in the capacitor after inserting the dielectric?

1) 5.3×10⁻¹⁰J
2) 4.0×10−⁻¹⁰J
3) 4.6×10⁻¹⁰J
4) 5.6×10⁻¹⁰J

Option 1

Option 2

Option 3 Solution:

Solution

$$C=10pF=10 \times 10^{-12} F$$

Voltage= $$Q=4.0V$$

Dielectric slab $$(K=5.0)$$

Electric potential energy= $$K.q\cfrac {1}{2}eV2^2$$

$$=\cfrac {1}{2}\times 4\times 10\times 10^{-12}$$

$$=20 \times 10^{-12} J$$

Now, after inserting dielectric, energy available in the capacitor

$$=\cfrac {1}{2}\times KC\times \left(\cfrac {V}{K}\right)^2$$

$$=\cfrac {1}{2}\times 5\times 10\times 10^{-12}\times \cfrac {4 \times 4}{5 \times 5}$$

$$=16 \times 10^{-12}$$ .

Total electric potential= $$(20+16)\times 10^{-12}$$

$$=46 \times 10^{-12}$$

$$=4.6\times 10^{-10} J$$ .

Option 4

2 Kinematics

How many seconds does it take a metal ball to drop 200. m from rest?
1) 6.39 s
2) 4.28 s
3) 8.58 s
4) 7.29 s

Option 1 Solution:

Option 2

Option 3

Option 4

3 Laws of Motion

A cart of mass has a block of mass m attached to it as shown in figure. The coefficient of friction between the block and the cart is µ. What is the minimum acceleration of the cart so that the block m does not fall

1)µg
2) g/µ
3)µ/g
4)Mµg/m

Option 1

Option 2 Solution:

Solution

Let a be the acceleration of trolley

From the FBD of block
N=ma
∴ µN=mg
µma=mg
a=g/µ

Option 3

Option 4

4 Motion of System of Particles and Rigid Body

A ball is whirled around in a horizontal circle at constant speed v by a string of length r. How much work is done by the string tension T during one complete revolution?

1. Zero

2. T·(2pr)

3. T·v

4. T· pr²

Option 1 Solution:

Option 2

Option 3

Option 4

5 Alternating Currents

An AC source of angular frequency ωω is fed across a resistor R and a capacitor C in series. The current registered is I. If now the frequency of source is changed to ω/3 (but maintaining the same voltage), the current in the circuit is found to be halved. The ratio of reactance to resistance at the original frequency ω is

1) $$\sqrt{\dfrac{3}{5}}$$
2) $$\sqrt{\dfrac{5}{3}}$$
3) $$\dfrac{3}{5}$$
4) $$\dfrac{5}{3}$$

Option 1 Solution:

Solution

$$I=\dfrac{V}{\sqrt{R^2+(\dfrac{1}{wc})^2}}$$, for w

for $$\dfrac{w}{3}, \dfrac{I}{2}=\dfrac{V}{\sqrt{R^2+(\dfrac{3}{wc})^2}}$$$$\dfrac{2}{\sqrt{R^2+\dfrac{9}{w^2C^2}}}=\dfrac{1}{\sqrt{R^2+\frac{1}{w^2C^2}}}$$$$4R^2+\dfrac{4}{w^2c^2}=R^2+\dfrac{9}{w^2c^2}$$

$$3R^2=\dfrac{5}{w^2c^2}$$

$$3R^2=5(x_c)^2$$ [at w]

$$\sqrt{\dfrac{3}{5}}=\dfrac{x_c}{R}$$

Option 2

Option 3

Option 4

6 Properties of Bulk Matter

Two soap bubbles A and B are kept in a closed chamber where the air is maintained at pressure 8N/m². The radii of bubbles A and B are 2 cm and 4 cm, respectively. Surface tension of the soap-water used to make bubbles is 0.04 N/m. Find the ratio nB/nA, where nA and nB are the number of moles of air in bubbles Aand B, respectively. [Neglect the effect of gravity.]

1) 2
2) 4
3) 6
4) 8

Option 1

Option 2

Option 3 Solution:

Solution

Excess pressure for soap bubble, $$\Delta P=\dfrac{4T}{R}$$ where $$T=$$ surface tension.
Here, $$\displaystyle P_A=P_0+\frac{4T}{R_A}=8+\frac{4(0.04)}{0.02}=16 N/m^2$$
and $$\displaystyle P_B=P_0+\frac{4T}{R_B}=8+\frac{4(0.04)}{0.04}=12 N/m^2$$
As $$PV=nRt , n\propto PV$$
Thus, $$\displaystyle \frac{n_A}{n_B}=\frac{P_AV_A}{P_BV_B}=\frac{16\times (4/3)\pi(0.02)^3}{12\times (4/3)\pi(0.04)^3}=1/6$$
$$\displaystyle \therefore \frac{n_B}{n_A}=6$$

Option 4

7 Motion of System of Particles and Rigid Body

A uniform wire of length l and mass m is bent in the form of a rectangle ABCD with AB=2BC . The moment inertia of this frame about BC is :

1) $$ \dfrac {11}{252} ml^2 $$
2) $$ \dfrac {8}{203} ml^2 $$
3) $$ \dfrac {5}{136} ml^2 $$
4) $$ \dfrac {7}{162} ml^2 $$

Option 1

Option 2

Option 3

Option 4 Solution:

Solution

The perimeter of the rectangle is given as,

$$l = 2\left( {AB + BC} \right)$$

$$l = 2\left( {2BC + BC} \right)$$

$$BC = AD = \frac{l}{6}$$

$$AB = DC = \frac{l}{3}$$

The mass of the length $$AB$$ is given as,

$${m_{AB}} = \frac{m}{3}$$

The mass of the length $$DC$$ is given as,

$${m_{DC}} = \frac{m}{3}$$

The mass of the length $$AD$$ is given as,

$${m_{AD}} = \frac{m}{6}$$

The total moment of inertia about the side $$BC$$ is given as,

$$M.I. = M.I{._{AB}} + M.I{._{DC}} + M.I{._{AD}}$$

$$M.I. = \frac{{m{l^2}}}{{81}} + \frac{{m{l^2}}}{{81}} + \frac{{m{l^2}}}{{27}}$$

$$M.I. = \frac{{7m{l^2}}}{{162}}$$

8 Motion of System of Particles and Rigid Body

A cube is in equilibrium on an inclined plane forming an angle of 30⁰ with the horizontal. If the distance of line of action of the normal reaction of the plane on the cube from the centre is a / 2√x. Find x

1) 4
2) 8
3) 3
4) 2

Option 1

Option 2

Option 3 Solution:

Solution

Net force perpendicular to the plane of the incline is zero,

$$\therefore$$ $$ mg cos(30^{\circ})$$ = N

Now consider the point on the centre of the base of the cube (on the incline). Frictional force passes through this point hence produces zero torque. Net torque about this point is zero. Only weight and normal produce a torque about this point.

$$\therefore$$ Writing torque equation we get,

$$mg sin(30^{\circ})\dfrac{a}{2}$$ = $$N(\dfrac{a}{2\sqrt{x}})$$

Substituting value of N,

$$mg sin(30^{\circ})\dfrac{a}{2}$$ = $$mg cos(30^{\circ})(\dfrac{a}{2\sqrt{x}})$$

$$tan(30^{\circ})$$ = $$\dfrac{1}{\sqrt{x}}$$

$$\therefore x = 3$$

Option 4

9 Alternating Currents

A coil, a capacitor and an AC source of rms voltage 24V are connected in series. By varying the frequency of the source, a maximum rms current of 6A is observed. If coil is connected to a dc battery of emf 12 volt and internal resistance 4Ω, then current through it in steady state is :

1) 2.4A
2) 1.8A
3) 1.5A
4) 1.2A

Option 1

Option 2

Option 3 Solution:

Solution

$$\displaystyle I_{rms}=\frac{V}{R}$$ (at resonance)
$$\displaystyle 6=\frac{24}{R}$$
$$\displaystyle \therefore R=4\Omega$$
$$\displaystyle I_{dc}=\frac{V}{R+r}=\frac{12}{4+4}=1.5A$$

Option 4

10 Electrostatics

n identical mercury droplets charged to the same potential V coalesce to form a single bigger drop. The potential of the new drop =

1) nn

2) n^2/3 n

3) n/n

4) n^1/3 n

Option 1

Option 2 Solution:

Solution

Suppose we have n identical drops each having radius r, capacitance C, charge q and potential V. If these drops are combined to form a big drop of radius R, capacitance C', charge Q and potential V will become : Charge on big drop

Q=nq Capacitance of big drop

C′ = n^1/3C

Hence potential of big drop V′=Q / C′ =nq / n1^/3C

⁼n ^2/3C

Option 3

Option 4

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