Q 1 :    

A box of mass 5 kg is pulled by a cord, up along a frictionless plane inclined at 30° with the horizontal. The tension in the cord is 30 N. The acceleration of the box is (Take g = 10 ms-2)                 [2024]
 

  • ms-2

     

  • zero

     

  • 0.1 ms-2

     

  • ms-2

     

(4)

From the given figure,

[IMAGE 34]----------------------------------

ma=T-mgsinθ

5a=30-5×10×12

5a=30-25

a=1m/s2



Q 2 :    

A block of mass 2 kg is placed on an inclined rough surface AC (as shown in figure) of coefficient of friction μ. If g = 10 ms-2, the net force (in N) on the block will be    [2023]

[IMAGE 35]
 

  • 103

     

  • zero

     

  • 10

     

  • 20

     

(2)

The given parameters are, Coefficient of friction, μ=13 and angle of friction is 30°.

[IMAGE 36]---------------------------------

Fnet=mgsinθ-μmgcosθ

         =mgsin30°-μmgcos30°

           =12mg-13mg32=12mg-12mg=0



Q 3 :    

A small block slides down on a smooth inclined plane, starting from rest at time t=0. Let Sn be the distance travelled by the block in the interval t=n-1 to t=n. Then, the ratio SnSn+1 is                  [2021]
 

  • 2n2n-1

     

  • 2n-12n

     

  • 2n-12n+1

     

  • 2n+12n-1

     

(3)

The acceleration is a=gsinθ

Initial velocity, u=0

Distance travelled in nth second Snth=u+12a(2n-1)

            Snth=12a(2n-1)                                                  ...(i)

            S(n+1)th=12a[2(n+1)-1]                                 ...(ii)

On dividing eqn (i) by (ii), we get

   SnthS(n+1)th=(2n-1)(2n+1)



Q 4 :    

Two bodies of mass 4 kg and 6 kg are tied to the ends of a massless string. The string passes over a pulley which is frictionless (see figure). The acceleration of the system in terms of acceleration due to gravity (g) is             [2020]

[IMAGE 37]
 

  • g

     

  • g/2

     

  • g/5

     

  • g/10

     

(3)

[IMAGE 38]----------------------------

Given : m1=4kg,m2=6kg

From the diagram,

T-m1g=m1a                     ...(i)

m2g-T=m2a                     ...(ii)

Solving equation (i) and (ii)

a=(m2-m1)gm2+m1=(6-4)g10=210g=g5



Q 5 :    

A block of mass m is placed on a smooth inclined wedge ABC of inclination θ as shown in the figure. The wedge is given an acceleration a towards the right. The relation between a and θ for the block to remain stationary on the wedge is                 [2018]

[IMAGE 39]
 

  • a=gcosecθ

     

  • a=gsinθ

     

  • a=gcosθ

     

  • a=gtanθ

     

(4)

[IMAGE 40]

In non-inertial frame,

Nsinθ=ma                                ...(i)

Ncosθ=mg                               ...(ii)

From (i) and (ii),

         tanθ=ag

a=gtanθ



Q 6 :    

Two blocks A and B of masses 3m and m respectively are connected by a massless and inextensible string. The whole system is suspended by a massless spring as shown in figure. The magnitudes of acceleration of A and B immediately after the string is cut, are respectively             [2017]

[IMAGE 41]
 

  • g3,g

     

  • g,g

     

  • g3,g3

     

  • g,g3

     

(1)

[IMAGE 42]

Before the string is cut

      kx=T+3mg                     ...(i)

     T=mg                                ...(ii)

From eqns. (i) and (ii)

     kx=4mg

Just after the string is cut

      T=0

aA=kx-3mg3m

aA=4mg-3mg3m=mg3m=g3

and aB=g



Q 7 :    

A balloon with mass m is descending down with an acceleration a (where a<g). How much mass should be removed from it so that it starts moving up with an acceleration a ?                     [2014]
 

  • 2mag+a

     

  • 2mag-a

     

  • mag+a

     

  • mag-a

     

(1)

Let F be the upthrust of the air. As the balloon is descending down with an acceleration a,  

  mg-F=ma                                                  ...(i)

[IMAGE 43]--------------------------------------------------

Let mass m0 be removed from the balloon so that it starts moving up with an acceleration a. Then,

           F-(m-m0)g=(m-m0)a

           F-mg+m0g=ma-m0a                                  ...(ii)

Adding equation (i) and equation (ii), we get

          m0g=2ma-m0a; m0g+m0a=2ma

          m0(g+a)=2ma

          m0=2maa+g