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 Multiple Choice QuestionsShort Answer Type

121.

A small spherical ball of mass m slides without friction from the top of a hemisphere of radius R. At what height will the ball lose contact with surface of the sphere ?


 Multiple Choice QuestionsMultiple Choice Questions

122.

Given A = 2 i^ + 3 j^ and B = i^ + j^. The component of vector A along vector B is

  • 12

  • 32

  • 52

  • 72


123.

f A = B + C have scalar magnitudes of 5, 4, 3 units respectively, then the angle between A and C is

  • cos-1 (3/5)

  • cos-1 (4/5)

  • π/2

  • sin-1 (3/4)


124.

A particle is projected from the ground with a kinetic energy E at an angle of 60° with the horizontal. Its kinetic energy at the highest point of its motion will be

  • E/2

  • E/2

  • E/4

  • E/8


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125.

A bullet on penetrating 30 cm into its target loses its velocity by 50%. What additional distance will it penetrate into the target before it comes to rest ?

  • 30 cm

  • 20 cm

  • 10 cm

  • 5 cm


126.

The velocity of a projectile at the initial point A is (2i + 3j) m/s. Its velocity (in m/s) at point B is

     

  • − 2i − 3j

  • − 2i + 3j

  • 2i − 3j

  • 2i + 3j


127.

A small object of uniform density rolls up a curved surface with an initial velocity v'. It reaches up to a maximum height of 3v24g with respect to the initial position. The object is

  • ring

  • solid sphere

  • hollow sphere

  • disc


128.

Consider three vectors A = i^ + j^ - 2k^B = i^ - j^ + k^ and C = 2i^ - 3j^ + 4k^. A vector X of the form αA + βB (α and β are numbers) is perpendicular to C. The ratio of α and β is

  • 1 : 1

  • 2 : 1

  • − 1 : 1

  • 3 : 1


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129.

A cricket ball thrown across a field is at heights h1 and h2 from the point of projection at times t1 and t2 respectively after the throw. The ball is caught by a fielder at the same height as that of projection. The time of flight of the hall in this journey is

  • h1t22 - h2t12h1t2 - h2t1

  • h1t12 + h2t22h2t1 + h1t2

  • h1t22 + h2t12h1t2 + h2t1

  • h1t12 - h2t22h1t1 - h2t2


A.

h1t22 - h2t12h1t2 - h2t1

For vertically moment

          h1 = u sin θ t1 - 12 gt12          (for h1)or  t1 = h1 + 12 gt12u sin θ                ...... (i)    h2 = u sin θ t2 - 12 gt22        (for h2)or  t2 = h2 + 12 gt22u sin θ              ....... (ii)Divide Eq. (i) by Eq. (ii)     t1t2 = h1 + 12 gt12 / u sin θh2 + 12 gt22 / u sin θ h1t2 - h2t1 = g2 t1t22 - t12t2The time of flight of the ball   T = 2usin θg = 2g u sin θ         [from Eq.(i)]      = 2g h1 + 12 gt12t1 = 2t1 h1g + t122      = h1t1 × 2g + t1 = h1t1 × t1t22 - t12t2h1t2 - h2t1 + t1      = h1t1t22 - h1t12t2 + h1t12t2 - h2t13t1 (h1t2 - h2t1)      = h1t22 - h2t12h1t2 - h2t1


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130.

A wooden block is floating on water kept in a beaker. 40% of the block is above the water surface. Now the beaker is kept inside a lift that starts going upward with acceleration equal to g/2. The block will then

  • sink

  • float with 10% above the water surface

  • float with 40% above the water surface

  • float with 70% above the water surface


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