If the gravitational force between two objects were proportional to 1/R (and not as 1/R2) where R is separation between them, then a particle in circular orbit under such a force would have its orbital speed v proportional to (a) …
Distance of geostationary satellite from the surface of earth radius (Re =6400 km )in terms of Re) is (a) 13.76 Re (b) 10.76 Re (c) 6.56 Re …
The distance between centre of the earth and moon is 384000 km. If the mass of the earth is 6 ×1024 kg and G = 6.66 × 10-11Nm2/kg2. The speed of the moon is nearly (a) 1 km/sec (b) …
Given radius of Earth ‘R’ and length of a day ‘T’ the height of a geostationary satellite is [G–Gravitational Constant, M–Mass of Earth] Answer: c (9)
The orbital speed of an artificial satellite very close to the surface of the earth is V° . Then the orbital speed of another artificial satellite at a height equal to three times the radius of the earth is (a) …
On applying a stress of 20×108 N/m2 the length of a perfectly elastic wire is doubled. Its Young’s modulus will be (a) 40×108 N/m2 (b) 20×108 N/m2 (c) 10×108 N/m2 (d) 5×108 N/m2 Answer-b (7)
How much force is required to produce an increase of 0.2% in the length of a brass wire of diameter 0.6 mm (Young’s modulus for brass =0.9×1011 N/m2 ) (a) Nearly 17 N …
A 5 m long aluminium wire ( Y=7×1010 N/m2) of diameter 3 mm supports a 40 kg mass. In order to have the same elongation in a copper wire ( Y=12×1010 N/m2) of the same length under the same weight, the …
Two similar wires under the same load yield elongation of 0.1 mm and 0.05 mm respectively. If the area of cross- section of the first wire is 4mm2 then the area of cross section of the second wire is (a) …
A steel wire of lm long and cross section area is 1mm2 hang from rigid end. When weight of 1kg is hung from it then change in length will be (given Y=2×1011N/m2) (a) 0.5 mm (b) 0.25 mm (c) …
Three rings each of mass M and radius R are arranged as shown in the figure. The moment of inertia of the system about YY’ will be (a) 3 MR2 (b) 3/2 MR2 (c) 5 MR2 (d) 7/2 MR2 Ans. d …
Four thin rods of same mass M and same length l, form a square as shown in figure. Moment of inertia of this system about an axis through centre O and perpendicular to its plane is (a) 4/3 Ml2 (b) …
A circular disc X of radius R is made from an iron plate of thickness t, and another disc Y of radius 4R is made from an iron plate of thickness t/4 . Then the relation between the moment of …
Two satellites of masses m1 and m2( m1>m2 ) are revolving round the earth in circular orbits of radius r1 and r2( r1>r2 ) Which of the following statements is true regarding their speeds v1 and v2 ? (a) v1 …
A rubber pipe of density 1..5×103 N/m2 and Young’s modulus is 5×106 N/m2 suspended from the roof. The length of the pipe is 8 m. What will be the change in length due to its own weight (a) 9.6 m …
A rod is fixed between two points at 20°C. The coefficient of linear expansion of material of rod is 1.1×10-5/°C and Young’s modulus is 1.2×1011 N/m2 . Find the stress developed in the rod if temperature of rod becomes 10°C …
The coefficient of linear expansion of brass and steel are α1 and α2 . If we take a brass rod of length l1 and steel rod of length l2 at 0°C, their difference in length (l2-l1) will remain the same at …
A wire of cross-sectional area is 3mm2 first stretched between two fixed points at a temperature of 20°C. Determine the tension when the temperature falls to 10°C. Coefficient of linear expansion α=10-5 °C-1 and Y=2×1011 N/m2 (a) 20 N (b) 30 …
3 particles each of mass m are kept at vertices of an equilateral triangle of side L. The gravitational field at centre due to these particles is (a) Zero (b) 3GM/L2 (c) 9 GM/L2 (d) ( 12/√3) (GM/L2) Answer: …
The relation between λ,ηand K for a elastic material is (a) (1/η)=(1/3λ)+(1/9k) (b) (1/k)=(1/3λ)+(1/9η) (c) (1/λ)=(1/3k)+(1/9η) …
The ratio of two specific heats of gas CP/CV for argon is 1.6 and for hydrogen is 1.4. Adiabatic elasticity of argon at pressure P is E. Adiabatic elasticity of hydrogen will also be equal to E at the pressure …
The string of a simple pendulum is replaced by a uniform rod of length L and mass M. If the mass of the bob of the pendulum is m, then for small oscillations its time period would be (assume radius …
A particle of mass 10 g is kept on the surface of a uniform sphere of mass 100 kg and radius 10 cm. Find the work to be done against the gravitational force between them to take the particle far …
A force of 103 newton stretches the length of a hanging wire by 1 millimetre. The force required to stretch a wire of same material and length but having four times the diameter by 1 millimetre is (a) 4×103 N …
The escape velocity from earth is ves. A body is projected with velocity 2 ves with what constant velocity will it move in the inter planetary space (a) ves (b) 3ves (c) √3 …
A planet has twice the radius but the mean density is 1/4 th as compared to earth. What is the ratio of escape velocity from earth to that from the planet (a) 3 : 1 …
The acceleration due to gravity on a planet is same as that on earth and its radius is four times that of earth. What will be the value of escape velocity on that planet if it is ve on earth …
The escape velocity for a body of mass 1 kg from the earth surface is 11.2 kms-1 The escape velocity for a body of mass 100 kg would be (a) 11.2 × 102 kms-2 …
The escape velocity for the earth is ve . The escape velocity for a planet whose radius is four times and density is nine times that of the earth, is (a) 36 ve (b) 12 ve (c) 6 ve (d) 20 ve Answer: b …
The radius of a planet is 1/4 of earth’s radius and its acceleration due to gravity is double that of earth’s acceleration due to gravity. How many times will the escape velocity at the planet’s surface be as compared to …
The escape velocity of a body on an imaginary planet which is thrice the radius of the earth and double the mass of the earth is ( ve is the escape velocity of earth) (a) √2/3 ve …
A mass of 6 × 1024 kg is to be compressed in a sphere in such a way that the escape velocity from the sphere is 3 × 108 m/s . Radius of the sphere should be (G = 6.67×10-11 N-m2/kg2) …
Two wires of the same material have lengths in the ratio 1 : 2 and their radii are in the ratio 1:√2 . If they are stretched by applying equal forces, the increase in their lengths will be in the …
A steel ring of radius r and cross-section area ‘A’ is fitted on to a wooden disc of radius R(R>r). If Young’s modulus be E, then the force with which the steel ring is expanded is (a) AE(R/r) …
The force required to stretch a steel wire of cross-section 1 cm2 to 1.1 times its length would be(Y=2×1011 N/m2) (a) 2×106 N (b) 2×103 N (c) 2×10-6 N …
In steel, the Young’s modulus and the strain at the breaking point are 2×1011 N/m2 and 0.15 respectively. The stress at the breaking point for steel is therefore (a) 1.33×1011 N/m2 (b) 1.33×1012 N/m2 (c) 7.5×10–13 N/m2 …
Poly – β – hydroxybutyrate – co – β hydroxyvalerate is (PHBV) is a copolymer of a. 3 – hydroxbutanoic and 2 – hydroxypentanoic acid b. 3 – hydroxbutanoic and 4 – hydroxypentanoic acid c. 2 – hydroxbutanoic and 3 …
The ratio of the radii of planets A and B is k1 and ratio of acceleration due to gravity on them is k2. The ratio of escape velocities from them will be (a) k1 k2 …
How many times is escape velocity ( Ve ) , of orbital velocity (V0 )for a satellite revolving near earth (a) √2 times (b) 2 times (c) 3 times (d) 4 times Answer: a (8)