The acceleration due to gravity on the earth’s surface at the poles is g and angular velocity of the earth about the axis passing through the pole is $$\omega$$. An object is weighed at the equator and at a height h above the poles by using a spring balance. If the weights are found to be same, then h is (h << R, where R is the radius of the earth)

A satellite is in an elliptical orbit around a planet P. It is observed that the velocity of the satellite when it is farthest from the planet is 6 times less than that when it is closest to the planet. The ratio of distances between the satellite and the planet at closest and farthest points is:

Two planets have masses M and 16 M and their radii are $$a$$ and 2$$a$$, respectively. The separation between the centres of the planets is 10$$a$$. A body of mass m is fired from the surface of the larger planet towards the smaller planet along the line joining their centres. For the body to be able to reach at the surface of smaller planet, the minimum firing speed needed is :

A body A of mass m is moving in a circular orbit of radius R about a planet. Another body B of mass {m \over 2} collides with A with a velocity which is half \left( {{{\overrightarrow v } \over 2}} \right) the instantaneous velocity$$\overrightarrow{v}$$ of A. The collision is completely inelastic. Then, the combined body :

The mass density of a spherical galaxy varies as {K \over r} over a large distance ‘r’ from its centre. In that region, a small star is in a circular orbit of radius R. Then the period of revolution, T depends on R as :

The height ‘h’ at which the weight of a body will be the same as that at the same depth ‘h’ from the surface of the earth is (Radius of the earth is R and effect of the rotation of the earth is neglected)

A satellite of mass m is launched vertically upwards with an initial speed u from the surface of the earth. After it reaches height R (R = radius of the earth), it ejects a rocket of mass {m \over {10}} so that subsequently the satellite moves in a circular orbit. The kinetic energy of the rocket is (G is the gravitational constant; M is the mass of the earth) :

A box weight 196 N on a spring balance at the north pole. Its weight recorded on the same balance if it is shifted to the equator is close to (Take g = 10 ms–2 at the north pole and the radius of the earth = 6400 km) :

Consider two solid spheres of radii R1 = 1m, R2 = 2m and masses M1 and M2, respectively. The gravitational field due to sphere (1) and (2) are shown. The value of {{{M_1}} \over {{M_2}}} is :

Planet A has mass M and radius R. Planet B has half the mass and half the radius of Planet A. If the escape velocities from the Planets A and B are vA and vB, respectively, then {{{v_A}} \over {{v_B}}} = {n \over 4}. The value of n is :

In the figure below, P and Q are two equally intense coherent sources emitting radiation of wavelength 20 m. The separation between P and Q is 5 m and the phase of P is ahead of that of Q by 90o. A, B and C are three distinct points of observation, each equidistant from the midpoint of PQ. The intensities of radiation at A, B, C will be in the ratio :

Two coherent sources of sound, S1 and S2, produce sound waves of the same wavelength, $$\lambda$$ = 1 m, in phase. S1 and S2 are placed 1.5 m apart (see fig). A listener, located at L, directly in front of S2 finds that the intensity is at a minimum when he is 2 m away from S2. The listener moves away from S1, keeping his distance from S2 fixed. The adjacent maximum of intensity is observed when the listener is at a distance d from S1. Then, d is :

Two light waves having the same wavelength $$\lambda$$ in vacuum are in phase initially. Then the first wave travels a path L1 through a medium of refractive index n1 while the second wave travels a path of length L2 through a medium of refractive index n2 . After this the phase difference between the two waves is :

In a Young’s double slit experiment, light of 500 nm is used to produce an interference pattern. When the distance between the slits is 0.05 mm, the angular width (in degree) of the fringes formed on the distance screen is close to

In a Young’s double slit experiment, 16 fringes are observed in a certain segment of the screen when light of a wavelength 700 nm is used. If the wavelength of light is changed to 400 nm, the number of fringes observed in the same segment of the screen would be

Interference fringes are observed on a screen by illuminating two thin slits 1 mm apart with a light source ($$\lambda$$ = 632.8 nm). The distance between the screen and the slits is 100 cm. If a bright fringe is observed on a screen at a distance of 1.27 mm from the central bright fringe, then the path difference between the waves, which are reaching this point from the slits is close is

A beam of plane polarised light of large cross-sectional area and uniform intensity of 3.3 Wm-2 falls normally on a polariser (cross sectional area 3 $$\times$$ 10-4 m2) which rotates about its axis with an angular speed of 31.4 rad/s. The energy of light passing through the polariser per revolution, is close to :

In a double slit experiment, at a certain point on the screen the path difference between the two interfering waves is {1 \over 8}th of a wavelength. The ratio of the intensity of light at that point to that at the centre of a bright fringe is :

In a Young's double slit experiment, the separation between the slits is 0.15 mm. in the experiment, a source of light of wavelengh 589 nm is used and the interference pattern is observed on a screen kept 1.5 m away. The separation between the successive bright fringes on the screen is :

Visible light of wavelength 6000 $$\times$$ 10-8 cm falls normally on a single slit and produces a diffraction pattern. It is found that the second diffraction minimum is at 60o from the central maximum. If the first minimum is produced at $$\theta$$1, then $$\theta$$1, is close to :