NEET physics

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Question
CBSEENPH11020679

A block of mass m is in contact with the cart C as shown in the figure


The coefficient of static friction between the block and the cart is μ. The acceleration α of the cart that will prevent the block from falling satisfies

  • straight alpha space greater than mg over straight mu
  • straight alpha space greater than space straight g over μm
  • straight alpha space greater or equal than space straight g over straight mu
  • straight alpha space less than space straight g over straight mu

Solution

C.

straight alpha space greater or equal than space straight g over straight mu

When a cart moves with some acceleration toward right then a pseudo force (mα) acts on block towards left. The force (mα) is action force by a block on the cart. Now, block will remain static w.r.t cart if frictional force 
μR space greater or equal than space mg

rightwards double arrow space straight mu space mα space greater or equal than space mg space space space left square bracket space as space straight R space equals space mα right square bracket

rightwards double arrow space straight alpha space greater or equal than space straight g over straight mu

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Question
CBSEENPH11020680

A circular disk of moment of inertia It is rotating in a horizontal plane, about its symmetry axis, with a constant angular speed.straight omega subscript straight i Another disk of moment of inertia Ib is dropped coaxially onto the rotating disk. Initially, the second disk has zero angular speed. Eventually, both the disks rotate with constant angular speed.straight omega subscript straight fThe energy lost by the initially rotating disc due to friction is 

  • 1 half fraction numerator straight I subscript straight b superscript 2 over denominator left parenthesis straight I subscript straight t plus straight I subscript straight b right parenthesis end fraction space straight omega subscript straight i superscript 2
  • 1 half fraction numerator straight I subscript straight t superscript 2 over denominator left parenthesis straight I subscript straight t plus straight I subscript straight b right parenthesis end fraction straight omega subscript straight i superscript 2
  • 1 half fraction numerator straight I subscript straight b minus straight I subscript straight t over denominator left parenthesis straight I subscript straight t plus straight I subscript straight b right parenthesis end fraction straight omega subscript straight i superscript 2
  • 1 half fraction numerator straight I subscript straight b straight I subscript straight t over denominator left parenthesis straight I subscript straight t plus straight I subscript straight b right parenthesis end fraction straight omega subscript straight i superscript 2

Solution

D.

1 half fraction numerator straight I subscript straight b straight I subscript straight t over denominator left parenthesis straight I subscript straight t plus straight I subscript straight b right parenthesis end fraction straight omega subscript straight i superscript 2

Loss of energy, 
increment straight E space equals space 1 half space straight I subscript straight t space straight omega subscript straight i superscript 2 minus 1 half fraction numerator straight I subscript straight t superscript 2 straight omega subscript straight i superscript 2 over denominator left parenthesis straight I subscript straight t plus straight I subscript straight b right parenthesis end fraction

equals space 1 half fraction numerator straight I subscript straight b straight I subscript straight t straight omega subscript straight i superscript 2 over denominator 2 left parenthesis straight I subscript straight t plus straight I subscript straight b right parenthesis end fraction

Question
CBSEENPH11020682

The radii of circular orbits of two satellites A and B of the earth are 4R and R, respectively. If the speed of satellite A is 3v. then the speed of satellite B will be

  • 3v/4

  • 6v

  • 12 v

  • 3v/2

Solution

B.

6v

Orbital velocity of satellite
straight v space equals space square root of GM over straight r end root
fraction numerator straight v subscript straight A over denominator straight v subscript straight B space end fraction space equals space square root of straight r subscript straight B over straight r subscript straight A end root space equals space square root of fraction numerator straight R over denominator 4 straight R end fraction end root space equals 1 half

therefore comma space straight v subscript straight A over straight v subscript straight B space equals space fraction numerator 3 straight v over denominator straight v subscript straight B end fraction space equals space 1 half

therefore comma space straight v subscript straight B space equals space 6 straight v

Question
CBSEENPH11020683

A ball is dropped from  a high-rise platform at t=0 starting from rest. After 6s another ball is thrown downwards from the same platform with a speed v. The two balls meet at t= 18s. What is the value of v? (take g= 10 ms-2)

  • 74 ms-2

  • 55 ms-1

  • 40 ms-1

  • 60 ms-1

Solution

A.

74 ms-2

For space first space ball space comma space straight u space equals space 0
therefore comma
straight s subscript 1 space equals space 1 half space gt subscript 1 superscript 2 space equals space 1 half space straight x space straight g space left parenthesis 18 right parenthesis squared
For space second space ball comma space initial space velocity space equals space straight v
therefore comma space
straight s subscript 2 space equals space vt subscript 2 space plus space 1 half space gt squared
straight t subscript 2 space equals space 18 minus 6 space equals space 12 space straight s
straight s subscript 2 space equals space straight v space straight x space 12 space plus space 1 half straight g space left parenthesis 12 right parenthesis squared
Here comma space space straight s subscript 1 space equals space straight s subscript 2
1 half straight g left parenthesis 18 right parenthesis squared space equals space 12 space straight v space plus space 1 half space left parenthesis straight g right parenthesis space left parenthesis 12 right parenthesis squared
straight v equals space 74 space ms to the power of negative 1 end exponent

Question
CBSEENPH11020686

A ball moving with velocity 2 ms-1 collides head on with another stationary ball of double the mass. If the coefficient of restitution is 0.5, then their velocities (in ms-1) after collision will be

  • 0,1

  • 1,1

  • 1,0.5

  • 0,2

Solution

A.

0,1

If two bodies collide head on with coefficient of restitution
straight e space equals space fraction numerator straight v subscript 2 minus straight v subscript 1 over denominator straight u subscript 1 minus straight u subscript 2 end fraction
From space the space law space of space conservation space of space linear space momentum

straight m subscript 1 straight u subscript 1 plus straight m subscript 2 straight u subscript 2 space equals space straight m subscript 1 straight v subscript 1 space plus straight m subscript 2 straight v subscript 2
rightwards double arrow space straight v subscript 1 space equals space open square brackets fraction numerator straight m subscript 1 minus em subscript 2 over denominator straight m subscript 1 plus straight m subscript 2 end fraction close square brackets straight u subscript 1 plus open square brackets fraction numerator left parenthesis 1 plus straight e right parenthesis straight m subscript 2 over denominator straight m subscript 1 plus straight m subscript 2 end fraction close square brackets straight u subscript 2
Substituting space straight u subscript 1 space equals space 2 space ms to the power of negative 1 end exponent comma space straight u subscript 2 space equals 0 comma space straight m subscript 1 equals straight m subscript 2 space and space straight m subscript 2 space equals space 2 straight m comma space straight e space equals space 0.5
we space get space space straight v subscript 1 space equals space open square brackets fraction numerator straight m minus straight m over denominator straight m plus 2 straight m end fraction close square brackets space straight x 2
similarly comma
straight v subscript 2 space equals open square brackets fraction numerator left parenthesis 1 plus straight e right parenthesis straight m subscript 1 over denominator straight m subscript 1 plus straight m subscript 2 end fraction close square brackets straight u subscript 1 space plus open square brackets fraction numerator straight m subscript 2 minus em subscript 1 over denominator straight m subscript 1 plus straight m subscript 2 end fraction close square brackets straight u subscript 2
equals space open square brackets fraction numerator 1.5 space straight x space straight m over denominator 3 straight m end fraction close square brackets space straight x space 2
space equals space 1 space ms to the power of negative 1 end exponent