JEE physics

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Question
CBSEENPH11020310

A screw gauge gives the following reading when used to measure the diameter of a wire. Main scale reading : 0 mm
Circular scale reading: 52 division
Given that 1 mm on the main scale corresponds to 100 divisions of the circular scale. The diameter of wire from the above data is

  • 0.052 cm

  • 0.026 cm

  • 0.005

  • 0.52

Solution

A.

0.052 cm

Least count of screw gauge =1/100 mm = 0.01 mn
Diameter - Divisions on cirular scale × least count + main scale reading = 52 × 1/100 + 0
100 = 0.52 mm
diameter = 0.052 cm

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Question
CBSEENPH11020311

A mass m hangs with the help of a string wrapped around a pulley on a frictionless bearing. The pulley has mass m and radius R.Assuming pulley to be a perfect uniform circular disc, the acceleration of the mass m, if the string does not slip on the pulley, is

  • g

  • 2/3g

  • g/3

  • 3/2g

Solution

B.

2/3g

WiredFaculty
mg - T = ma
WiredFaculty

Question
CBSEENPH11020312

Work done in increasing the size of a soap bubble from a radius of 3 cm to 5 cm is nearly. (Surface tension of soap solution = 0.03 Nm-1)

  • 4π mJ 

  • 0.2π mJ

  • 2π mJ

  • 0.4π mJ

Solution

D.

0.4π mJ

Work done = Change in surface energy
⇒ W = 2T x 4π (R22-R12)
 = 2 x 0.03 x 4π [ (5)2-(3)2] x 10-4
 = 0.4 π mJ

Question
CBSEENPH11020313

A thin horizontal circular disc is rotating about a vertical axis passing through its centre. An insect is at rest at a point near the rim of the disc. The insect now moves along a diameter of the disc to reach its other end. During the journey of the insect, the angular speed of the disc 

  • remains unchanged

  • continuously decreases

  • continuously increases 

  • first increases and then decreases

Solution

D.

first increases and then decreases

WiredFaculty
From angular momentum conservation about the vertical axis passing through centre. When the insect is coming from circumference to center. Moment of inertia first decreases then increase. So angular velocity increase than decrease. 

Question
CBSEENPH11020315

Two bodies of masses m and 4 m are placed at a distance r. The gravitational potential at a point on the line joining them where the gravitational field is zero is

  • -4Gm/r

  • -6Gm/r

  • -9Gm/r

  • zero

Solution

C.

-9Gm/r

WiredFaculty
WiredFaculty
= -3Gm/r - 6Gm/r = -9Gm/r