JEE PHYSICS question bank, MCQ , Notes, Lectures

1 Capacitance

The charge on the capacitor of capacitance C shown in the figure below will be

Option 1

Option 2

Option 3 Solution:

SOLUTION

Option 4

2 Rotational Motion

The moment of inertia of a sphere of mass M and radius R about an axis passing through its centre
is 2 / 5MR². The radius of gyration of the sphere about a parallel axis to the above and tangent to the sphere is

1) 7/5R 2) 3/5R 3) 4)

Option 1

Option 2

Option 3 Solution:

SOLUTION :
Moment of inertia of solid sphere about its COM Icom=2/5MR²
Moment of inertia of solid sphere about an axis parallel to COM and tangent to the sphere

Option 4

3 Work Power Energy

A spring of force constant 10 N per meter has an initial stretch 0.20 m. In changing the stretch to 0.25m, the increase in potential energy is about

1) 0.1 joule 2) 0.2 joule 3) 0.3 joule 4) 0.5 joule

Option 1 Solution:

SOLUTION:−

Elastic potential energy of spring = ½ kx²
Change in energy = ½10 ( 0.25² - 0.20²)
Change in energy = 0.1125 J

Option 2

Option 3

Option 4

4 Alternating Currents

A coil of inductance�5.0�mH�and negligible resistance is connected to an alternating voltage�V=10sin(100t). The peak current in the circuit will be:

1) 2�amp
2) 1�amp
3)�10�amp
4)�20�amp

�

Option 1

Option 2

Option 3

Option 4 Solution:

Solution

Inductive reactance $${ X }_{ L }=\omega L\\ =100\times 5\times { 10 }^{ -3 }=0.5\Omega $$

�

$${ V }_{ max }=Peak \ Voltage=10V$$

�

Peak Current $${ I }_{ p }=\dfrac { PeakVoltage }{ { X }_{ L } } =\dfrac { 10 }{ 0.5 } =20Amps$$

�

5 Semiconductors and Electronic Devices

What are the reading of the ammeters A1and A2 shown in figure . Neglect the resistance of the meters.

1) 0.1A, 0.2 A
2) 0.02 A, 0
3) 0.2 A, 0
4) 0, 0.2 A

Option 1

Option 2

Option 3

Option 4 Solution:

SOLUTION

6 Atoms and Nuclei

�A photon of energy 10.2 eV collide in elastically with hydrogen atom in ground state after few microseconds another photon of energy 15eV collide in elastically with same hydrogen atom finally by a suitable detector we find.

(1) Photon of energy 3.4 eV and electron of energy 1.4 eV.

(2) Photon of energy 10.2 eV and electron of energy 1.4 eV.

(3) Two Photons of energy 3.4eV

(4) Two Photons of energy 10.2 eV

Option 1

Option 2 Solution:

Option 3

Option 4

7 Magnetic Effects of Current

Yellow light of wavelength 6000Å produces fringes of width 0.8mm in Youngs double slit experiment. If the source is replaced by another monochromatic source of wavelength 7500Å and the separation between the slits is doubled then the fringe width becomes

1) 0.1mm
2) 0.5mm
3) 4.3mm
4) 1mm

Option 1

Option 2 Solution:

Solution:

Fringe width in first case, $$\beta_{1} = \frac{D\lambda_{1}}{d}\quad...\left(i\right) $$

Fringe width in second case, $$\beta_{2}=\frac{D\lambda_{2}}{2d} \quad...\left(ii\right) $$

Divide equation $$\left(ii\right)$$ by $$\left(i\right)$$,

$$ \therefore \frac{\beta_{2}}{\beta_{1}}= \frac{D\lambda_{2}/ 2d}{D\lambda_{1} /d} = \frac{1}{2} \cdot\frac{\lambda_{2}}{\lambda_{1}} $$ or

$$\beta_{2} = \frac{1}{2}\cdot\frac{\lambda_{2}}{\lambda_{1}} \cdot\beta_{1} $$

$$\therefore \beta_{2} = \frac{1}{2}\times\frac{ 7500 Å}{6000 Å} \times 0.8 mm $$

$$= 0.5 mm $$

Option 3

Option 4

8 Magnetic Effects of Current

The electric current in a circular coil of two turns produced a magnetic induction of $$0.2\, T$$ at its centre. The coil is unwound and then rewound into a circular coil of four turns. If same current flows in the coil, the magnetic induction at the centre of the coil now is

1) $$0.2\, T$$
2) $$0.4\, T$$
3) $$0.6\,T$$
4) $$0.8\, T$$

Option 1

Option 2

Option 3

Option 4 Solution:

Solution:

When there are two turns in the coil, then

$$l=2\times2\pi r_{1}$$ or $$r_{1}=\frac{l}{4\pi}$$

then $$B_{1}=\frac{\mu_{0} N_{1}I}{2r_{1}}=\frac{\mu_{0}\times2\times I}{2\times\left(l / 4\pi\right)}=\frac{\mu_{0}\,4\pi I}{l}$$

When there are four turns in the coil, then

$$l=4 \times2\pi r_{2}$$ or $$r_{2}=\frac{l}{8\pi}$$

Then $$B_{2}=\frac{\mu_{0}\, N_{2}I}{2r_{2}} = \frac{\mu_{0}\times4\times I}{2\times\left(l 8\pi\right)}=\frac{\mu_{0}\,16\pi I}{l}$$

$$\frac{B_{1}}{B_{2}}=\frac{4}{16}=\frac{1}{4}$$ or $$B_{2}=4B_{1}=4\times0.2\,T$$

$$=0.8\, T $$

9 Magnetic Effects of Current

A metal disc of radius 100 cm is rotated at a constant angular speed of 60 rads ^-1 in a plane at right angles to an external field of magnetic induction 0.05 Wbm^-2. The end' induced between the centre and a point on the rim will be

1) 3 V
2) 1.5 V
3) 6V
4) 9 V

Option 1

Option 2 Solution:

SOLUTION

Option 3

Option 4

10 Wave Optics

Interference fringes were produced in Youngs double slit experiment using light of wavelength $$5000 Å$$. When a film of material $$2.5 \times 10^{-3}\, cm$$ thick was placed over one of the slits, the fringe pattern shifted by a distance equal to $$20$$ fringe widths. The refractive index of the material of the film is

1) $$1.25$$
2) $$1.33$$
3) $$1.4$$
4) $$1.5$$

Option 1

Option 2

Option 3 Solution:

Solution:

Fringe width, $$\beta = \frac{\lambda D}{d} \quad ...(i)$$

where $$D$$ is the distance between the screen and slit and $$d$$ is the distance between two slits.

When a film of thickness $$t$$ and refractive index $$\mu$$ is placed over one of the slit, the fringe pattern is shifted by distance $$S$$ and is given by

$$S = \frac{(\mu - 1) t D}{d} \quad ...(i)$$

Given : $$S = 20 \beta \quad ...(iii)$$

From equations $$(i), (ii)$$ and $$(iii)$$ we get,

$$(\mu -1) t = 20 \lambda $$

or $$(\mu -1) = \frac{20\lambda}{t}$$

$$= \frac{20\times 5000 \times 10^{-8}\,cm}{2.5 \times 10^{-3}\,cm}$$

$$\mu - 1 = 0.4 $$ or $$\mu = 1.4$$

Option 4

JEE PHYSICS

DAILY PHYSICS TEST QUESTIONS

Question Bank

Solution

MOCK TEST