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Water rises to a height of 10cm in a capillary tube and mercury falls to a depth of 3.42 cm in the same capillary tube . If the density of mercury is 13.6 g/cc and the angle of contact of mercury and water are 135° and 0° respectively , the ratio of surface tension of water to mercury is : A) 1: 0.15 B) 1:3 C) 1: 6.5 D) 1.5: 1

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A planet of radius R has an acceleration due to gravity of 𝑔_𝑠 on its surface . A deep smooth tunnel is dug on this planet , radially inward , to reach a point P located at a distance of 𝑅/2 from the centre of the planet . Assume that the planet has uniform density . the kinetic energy required to be given to a small body of mass m , projected radially outward from P , sop that it gains a maximum altitude equal to the thrice the radius of the planet from its surface is equal to 63/16m𝑔_𝑠R 3/8m𝑔_𝑠R 9/8m𝑔_𝑠R 21/8m𝑔_𝑠R

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From a circular disc of radius R and mass 9M , a small disc of radius R/3 is removed from the disc . The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through o is 4M𝑅^2 40/9M𝑅^2 10M𝑅^2 37/9M𝑅^2

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particle is moving parallel to x-axis as shown in the figure such that all instants the y – component of its position vector is constant and equal to ‘b ‘ . The angular velocity of the particle p about origin at the given instant is A) 𝑉/𝑏 cos⁡𝜃 B) 𝑉/𝑏 sin⁡𝜃 C) 𝑉/𝑏 sin^2⁡𝜃 D) vb

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A system of two bodies of masses ‘m’ and M being interconnected by a spring of stiffness k , in its natural length , moves towards a rigid wall on a smooth horizontal surface as shown in figure with a K.E. system ‘E’ . If the body M sticks to the wall after the collision , the maximum compression of the springs will be : A) √(𝑚𝐸/𝑀𝑘) B) √(2𝑚𝐸/(𝑀+𝑚)𝑘) C) √((2(𝑚+𝑀)𝐸)/𝑘(𝑚) ) D) √(2𝑀𝐸/(𝑀+𝑚)𝑘)

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An object is placed on the surface of a smooth inclined plane of inclination 𝜃 . It takes time t to reach the bottom . If the same object is allowed to slide down a rough inclined plane of same inclination 𝜃 , so as to move the same distance , it takes time ‘nt’ to reach the bottom where n is a number greater than 1 . The coefficient of friction 𝜇 𝑖𝑠 𝑔𝑖𝑣𝑒𝑛 𝑏𝑦 : 𝜇 = tan⁡𝜃 (1 - 1/𝑛^2 ) 𝜇 = cot⁡𝜃 (1 - 1/𝑛^2 ) 𝜇 = tan⁡𝜃 〖"(1 - " 1/𝑛^2 ")" 〗^(1/2) 𝜇 = cot⁡𝜃 〖"(1 − " 1/𝑛^2 ")" 〗^(1/2)

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A sphere of mass 𝑚_1=2kg collides with a sphere of mass 𝑚_2=2kg which is at rest . Mass 𝑚_1 , after the collision will move at right angle to the joining centers of two spheres at the time of collision , assuming colliding surfaces are smooth , if the coefficient of restitution is

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Physics Electrostatics Class 12 Cbse Important questions last 5years questions and answer chapter 1

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Haloalkanes and haloarenes class 12 chemistry important questions for cbse examination

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class 8 chapter 7 science Ncert solution WORD FILE

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