JEE PHYSICS question bank, MCQ , Notes, Lectures

1 Kinematics

An ant goes from P to R on a circular path in 20 second Radius OP=10m. What is the average speed and average velocity of it ?

1) (π/6) m/s, 3 m/s 2) (π/3) m/s, √3/2 3) (π/3) m/s,√3 4) π √6

Option 1

Option 2 Solution:

Length of arc=(2π/3) × (op)
Speed=Length of arc/ time

Displacement PR

From figure angle POM=60°
thus PM=OPsin60
PM=10 (√3 /2)
Thus PR=2PM=10( √3)
Velocity=PR/time=√3/2

Option 3

Option 4

2 Electromagnetic Induction

A solenoid of length 30 cm with 10 turns per�centimetre and area of cross-Section 40�cm² completely surrounds another�co-axial solenoid of same length, area of Cross-section 20�cm^2�with 40 turns per centimetre. The�mutual inductance of the

1) 10 H
2) 8 H
3) 3mH
4)�30 mH

�

Option 1

Option 2

Option 3 Solution:

Solution

Mutual inductance of the system,
M=μ0n1n₂A₂l
Where�A₂is�the area of inner solenoid.
n1=10cm⁻¹=1000m⁻¹
n2=40cm⁻¹=4000m⁻¹
l=30cm=30�10m⁻¹
A2=20cm²=20�10⁻⁴m²
∴M=4π�10⁻⁷�1000�4000�20�10⁻⁴�30�10⁻²
=301.44�10⁻⁵H=3mH

Option 4

3 Electrostatic Potential

A spherical shell of radius $$10\, cm$$ is carrying a charge $$q$$. If the electric potential at distances $$5 \, cm, 10 \, cm$$ and $$15 \, cm$$ from the centre of the spherical shell is $$V_1, V_2$ and $V_3$ respectively, then respectively, then

1) $$V_1>V_2>V_3$$
2) $$V_1$$<$$V_2$$<$$V_3$$
3) $$V_1=V_2>V_3$$
4) $$V_1=$$$$V_2$$<$$V_3$$

Option 1

Option 2

Option 3 Solution:

Solution:

Inside the spherical shell potential is same

$$\Rightarrow\quad\quad\quad V_{1}=V_{2}$$

$$\quad \quad \quad \quad V\propto \frac{1}{d}$$

i.e. $$ \quad \quad V_{1}=V_{2}>V_{3}$$

Option 4

4 Kinematics

A small particle of mass m is projected at an angle θ with the X–axis with an initial velocity v_o in the X–Y plane as shown in the figure. At a time

Option 1

Option 2

Option 3 Solution:

SOLUTION

Option 4

5 Gravitation

An object weights 72 N on the earth. Its weight at a height R/2 from earth is = .............. N

1) 72
2) 0
3) 56
4) 32

Option 1

Option 2

Option 3

Option 4 Solution:

SOLUTION

6 Laws of Motion

An object of mass m is hanging by a string from the ceiling of an elevator. The elevator is moving upward, but slowing down. What is the tension in the string?

1) Less than mg. 2) Exactly mg. 3) Greater than mg. 4) Zero.

Option 1 Solution:

Solution :
g_eff = g – a (going up & slowing down => acceleration downward)
=> T = mg_eff = m(g – a) < mg

Option 2

Option 3

Option 4

7 Dual Nature Of Radiation And Matter

The radiation corresponding to 3 -> 2 transition of hydrogen atom falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field of 3 x 10^-4T. If the radius of the largest circular path followed by these electrons is 10.0 mm, the work function of the metal is close to

1) 0.8eV
2) 1.6eV
3) 1.8eV
4) 1.1eV

Option 1

Option 2

Option 3

Option 4 Solution:

SOLUTION

8 Electrostatic Potential

1000 drops of water of radius 1 cm each carrying a charge of 10 esu combine to form a single drop. The capacitance increases thereby

1) 1 time
2) 10 times
3) 100 times
4) 1000 times

Option 1

Option 2 Solution:

Solution:

$$n^{\frac{1}{3}}r =R$$

$$n^{\frac{1}{3}}C\,_{small\,drop}\,^{=C\,_{bigger \,drop }}$$

$$\left(10\right)^{3\times\frac{1}{3}}C\,_{small\,drop}\,^{=C\,_{bigger\,drop}}$$

Option 3

Option 4

9 Magnetic Effects of Current and Magnetism

A current of 50 A is passed through a straight wire of length 6 cm,then the magnetic induction at a point 5 cm from the either end of the wire is (1 gauss =10⁻⁴ T)

1) 2.5 gauss
2) 1.25 gauss
3) 1.2 gauss
4) 3.0 gauss

Option 1

Option 2

Option 3 Solution:

Solution

The magnetic field due to the finite current carrying wire is:

$$B=\dfrac{\mu_0I}{4\pi r}[\sin\theta_1+\sin\theta_2]$$

$$=10^{-7} \times \dfrac{50}{5\times 10^{-2}}[\dfrac{3}{5}+\dfrac{3}{5}]=1.2 \times 10^{-4} T=1.2 $$ gauss

Option 4

10 Work Power Energy

A particle of mass $$m$$ moving in the $$x$$-direction with speed $$2v$$ is hit by another particle of mass $$2m$$ moving in the $$y$$-direction with speed $$v$$. If the collision is perfectly inelastic, the percentage loss in the energy during the collision is close to

1) 50 %
2) 56 %
3) 62 %
4) 44 %

Option 1

Option 2 Solution:

Solution:

In all type of collisions, momentum of the system always remains constant. In perfectly inelastic collision, particles stick together and move with a common velocity.

Let this velocity is $$v_c$$. Then, initial momentum of system = final momentum of system

or m ( 2 v) $$ \widehat{ i} + 2 m ( v) \widehat{j} = ( m + 2 m) \, v_c $$

$$\therefore v_c = \frac{ 2}{ 3} ( v \widehat{ i } + v \widehat{ j} )$$

$$ | v_c | \, or \, v_c \, or \, speed = \sqrt{ \bigg( \frac{2}{3} v \bigg)^2 + \bigg( \frac{ 2}{ 3} v \bigg)^2 } $$

= $$ \frac{ 2 \sqrt 2 }{ 3} v $$

Initial kinetic energy

$$ K_i = \frac{1}{2} ( m) \, ( 2 v)^2 + \frac{1}{2} (2 m) \, (v)^2 = 3 mv^2 $$

Final kinetic energy

$$ K_ f = \frac{1}{2} ( 3m) \bigg( \frac{ 2 \sqrt 2 }{ 3 } v \bigg)^2 = \frac{ 4 }{ 3} mv^2 $$

Fractional loss = $$ \bigg( \frac{ K_i - K_f }{ K_i } \bigg) \times 100 $$

= $$ \Bigg [ \frac{ ( 3mv^2 ) - \bigg( \frac{ 4 }{ 3} mv^2 \bigg) }{ ( mv^2 ) } \Bigg ] \times 100 = 56 %$$

Option 3

Option 4

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