**Mohammed **
**Asif**
**Name :**
**Roll No. :**
**Topic : **
**Ph : 9391326657, 64606657**

**Multiple choice questions (only one is correct)**

**1.** **A body of specific gravity 6, weighs 0.9kg when placed in one pan (say pan A) and 1.6kg when **

**placed on the other pan (pan B) of a false balance. The beam is horizontal when both the pans **
**are empty. Now if the body is suspended form pan A and fully immersed in water, it will weigh**

**a) 0.5kg ** **b) 0.6kg ** **c) 0.75kg** **d) 0.8kg**

**2.** **The pulley has mass M >m. String is massless. The above system is released from rest from the **
**position shown. Then **

** **

**a) Body (1) will slowly come down, till equilibrium is attained at level AA’ shown.**

**b) The system will perform oscillations with equilibrium position at level AA’ and amplitude **
** 0.5m**

**c) The response of the system depends on whether pulley-string interface has friction or not.**
**d) The system will continue to be in the same initial position.**

**3.** **In the position shown above Let T1 be the tension in the left part of the string, T2 = Tension in **
**the right part, N = Normal reaction on block (2) by the resting surface. Then**

** **
**a) T1 = T2 = 0 and N = mg** **b) T1 =mg, T2 = 0, N =mg**
**c) T1 = T2 = mg and N = 0** **d) T1 = T2 =**
2
*mg*
** and N=**
2
*mg*

**4.** **A body of mass m was slowly hauled up the hill and down the hill onto the other side by a force **
→

**F which at each point was directed along a tangent to the trajectory. The work performed by ****this force, if the coefficient of friction is **µ1**uphill and **µ2** dowbhill, is**

** **

**a) mg **

### (

µ11+µ22### )

**b) mg**

### (

2*h*+µ11 +µ22

### )

**c) mg **

### (

*h*+µ11 +µ22

### )

**d) mg**

### (

µ11−µ22### )

**5.** **The bob of a pendulum is taken to position A and given an initial velocity u in the direction **
**shown, in the following cases.**

** **

**If the minimum value of u so that the pendulum reaches position OB in case (i) and position OC **
**in case (ii) are u1 in case (i) & u2 in case (ii), then **

### (

2### )

1

2 _{is}_{g}_{=}_{10}* _{ms}*−

*u*
*u*

**6.** **A conveyor belt carrying powdery material is at an angle 370 _{ to the horizontal and moves at a }**

**constant speed of 1ms-1**

_{ as shown. Through a small hole in the belt, the powdery material drops }**down at a constant rate of 1kg per second. What is the force to be applied on the belt along the**

**direction of its motion so as to maintain its constant speed of 1ms-1**

_{?}** **

**a) -1 N** **b) Zero** **c) +1 N** **d) None of the above**

**7.** **Small body A on a hemispherical body B which is on a wedge C which is on a smooth horizontal **

**surface. System is released from rest from the position shown when **

*cm*
*cm*
*B* 10 , 5
,
370 _{=} _{=}
=
α **.**
**When ** 53 , *C*
0
=

α **is 4.5cm ***B***at that instant is (neglect friction)**

** **

**a) 9 cm** **b) 10 cm** **c) 10.5 cm** **d) 11 cm**

**8.** **A pendulum consists of a mass m attached to the end of a light string 0.5m long. It can oscillate **
**in the vertical plane. If it is let go in the horizontal position and has an inelastic collision with **
**the floor, e = 0.5, the rebound velocity is (ms-1 _{) (the angle turned is}**π

_{/}

_{6}

_{ ) (g = 10ms}-2_{)}** **
**a) **
2
5 ** _{b) }**
4
5

**c)**4 35

**4 45**

_{d) }**9.** **The track is in the vertical plane. The track is rough with friction coefficient **µ**. A particle at A **
**is allowed to slide down and goes upto B and returns. Heights of A and B are 0.2m and 0.1m **
**respectively, as shown. The maximum possible value of **µ** is**

** **
**a) **
4
1
**b) **
7
1
**c) **
6
1
**d) **
5
1

**10.** **A particle is projected form a point on a horizontal floor. After it has three collisions with the **
**floor, it is found that the ratio of maximum height to minimum peaks reached by it is **_{} 36 _{}

12 2 10

**. **
**The coefficient of restitution is**

**a) .5** **b) 0.64** **c) 0.8** **d) 0.9**

**11.** **The potential energy function along the positive x axis is given by **

### ( )

*x*
*b*
*ax*
*x*

*U* =− + **, a, b are **

**constants. If it is known that the system has only one stable equilibrium configuration, the **
**possible values of a and b are **

**a) a = 1, b = 2** **b) a = 1, b = -2** **c) a = -1, b = 2** **d) a = -1, b = -2**

**12.** **The acceleration of the 1kg block immediately after the string is cut is (g = 10ms-2 _{).}**

** **

**a) 4ms-2** _{b) 4.1ms}-2_{c) 16ms}-2_{d) 40ms}-2

**13.** **A circular pan with its side wall inclined inward at 530 _{ as shown is rotating about its central }**

**vertical axis with a constant angular velocity. A ball placed at the edge rotates along with it as**

**shown. If the ball exerts a force of 22.5 N on the side wall and 23.5N on the bottom surface, the**

**mass of the ball is**

** **

**a) 1kg** **b) 2kg** **c) 3kg** **4kg**

**14.** **Acceleration of 10kg block, when system released from rest, is **

** **

**a) 0.5ms-2** _{b) 1 ms}-2_{c) 1.25 ms}-2_{d) 1.66ms}-2

**15.** **Case(A)**

**Case (B) **

**The reaction force between the 5kg and 6kg block in case A and case B will be**

**a) equal and non zero** **b) unequal with case A being more**

**c) unequal with case being more** **d) equal and zero**

**For Question No. 16 to 20:**

**Each question consider of two statements: one is Assertion (A) and the other is Reason(R). You **
**are to examine these two statements and select the answer using the code given below.**

**a) Both A and R individually correct, but R is the correct explanation of A**
**b) Both A and R individually correct, but R is the not correct explanation of A**

**c) A is true but R is false** **d) A is false but R is true**

**16.** **Assertion (A): If a block is released form rest, when reaches the bottom point of the wedge, its **
** speed is same irrespective whether the wedge is fixed or the wedge is fee to move.**
**Reason(R): Mechanical energy is conserved in both cases.**

**17.** **Assertion (A): A body is at rest on floor. You lift it vertically up and bring it to rest at a point h **
** above the ground. The work done by you is zero.**

**Reason(R): Any non-zero work done by a force on a body results in change in kinetic energy of **
** the body.**

**18.** **Assertion (A): The negative of the work done by the conservative internal forces on s system **
** equals to change in its potential energy.**

**19.** **Assertion (A): In a tug of war that team wins which applies more tension force on string then **
** their opponents.**

**Reason(R): The winning team must be having stronger players.**

**20.** **Assertion (A): If a block starts moving at t = 0 with an acceleration ‘a’ then its kinetic energy at **
** time t with respect to two non-inertial frames which have same acceleration in a **
** direction is not always same.**

**Reason(R): The work energy theorem **∆*K* −∆*W* ** is not for non-inertial frame.**
**Match the following Questions 21 to 25**

**21.** **Two columns are given in each question. Match the elements of Column-I with Column-II.**

** Column-I** ** Column-II**

**i) Friction force** **p) Contact force**

**ii) Normal reaction** **q) Electromagnetic force**

**iii) Tension in a string ** **r) Gravitational force**

**iv) Force between two charges of mass m** **s) Nuclear force**

**a) (i-p,q), (ii-p,q), (iii-q,s), (iv-q,r)** **b) (i-p,q), (ii-p,q), (iii-q), (iv-q,r)**
**c) (i-p,q), (ii-p), (iii-q,s), (iv-q)** **d) (i-q), (ii-p), (iii-q,s), (iv-p,q,r)**

**22. ** **In Column-I there are some motions of a body and Column-II contains the list of concepts that **
**can be used for the analysis of these motions**

**Column-I** ** ** ** Column-II**

**i) A body moving in a vertical circle** **p) Conservation of energy**

**ii) A body moving in a horizontal circle** **q) Conservation of momentum**

**iii) A body dropped form a height on a block attached **

** to the top of a vertical spring** **r) Centripetal force**

**iv) Rocket propulsion** **s) Centrifugal force**

**a) (i-p,q,r), (ii-q,r,s), (iii-p,q,s), (iv-p)** **b) (i-p,s), (ii-p,r), (iii-q), (iv-p,q,r)**
**c) (i-p,r), (ii-q,s), (iii-p), (iv-p,q)** **d) (i-p,r,s), (ii-r,s), (iii-p,q), (iv-q)**

**23.** **When two bodies collide they come in contact at t = t1 and loses contact at t = t2. This (t2-t1) is a **
**very small time interval. Consider this small time interval and match the following.**

**i) Elastic collision ** **p) Kinetic energy decreases and potential **

** energy increases and then potential **

** energy decreases and kinetic energy increases**

**ii) Inelastic collision** **q) Kinetic energy + potential energy is conserved **

**iii) Perfectly inelastic collision** **r) Momentum is conserved**

**iv)Oblique elastic collision** **s) Kinetic energy decreases and potential energy **

** increases and then situation remains same**
**a) (i-p,q,r), (ii-q,r,s), (iii-p,q,s), (iv-p)** **b) (i-p,s), (ii-p,r), (iii-q), (iv-p,q,r)**

**c) (i-p,r), (ii-q,s), (iii-p), (iv-p,q)** **d) (i-p,r,s), (ii-r,s), (iii-p,q), (iv-q)**

**24.** **For each of the following four cases of Column-I match the range of force F in column B so that **
**the block m is not slipping on the surface with which it has contact.**

**For all cases m = 2kg; **µ=0.2** Take g = 10m/s2**
**Column-I** ** ** ** Column-II**
**p) (48N, 80N)**
**q) **
_{N}* _{N}*
11
500
,
19
500

**r)**

_{N}*11 380 , 19 220*

_{N}**s)**

_{N}*17 380 , 23 220*

_{N}**a) (i-p), (ii-q), (iii-q,r), (iv-p,q,r,s)** **b) (i-p), (ii-q), (iii-r,s), (iv-q,r)**
**c) (i-q), (ii-p,s), (iii-q,s), (iv-r,s)** **d) (i-q,r), (ii-p,q), (iii-r), (iv-p,q,r)**

**25.** **A body is moving in a vertical circle of radius R, considering motion from top point of circle **
**bottom most point of circle, match the following.**

** Column-I** ** Column-II**

**i) Tangential accelertaion** **p) Always increases**

**ii) Centripetal acceleration** **q) Always decreases **

**iii) Angular velocity** **r) First increases then decreases**

**iv) Potential energy** **s) Variable depending on angle made with **

** vertical by string**

**t) Variable independent of the made angle with **
** vertical by string **

**a) (i-r,s), (ii-q,r), (iii-p,t), (iv-p,q)** **b) (i-q,s), (ii-r,s), (iii-p,s), (iv-q,r)**
**c) (i-p,s), (ii-q,r), (iii-q,s), (iv-p,r)** **d) (i-r,s), (ii-q,s), (iii-p,s), (iv-p,s)**
**Write the final answer to each question in this section in the column provided.**

**26.** **The block ABCDE of mass 5m has BC part spherical, of radius 1m, is on frictionless horizontal **
**surface. BC is quarter circle. A small mass m is released at B and slides down. How far away **
**form D does it hit the floor? (g = 10ms-2 _{)}**

** **

**27.** **The kinetic energy of a particle of mass 1kg moving along a circle of radius 1m depends on time **
**t as K = t4 _{. Fid the force acting on the particle as function of t.}**

**28.** **Spring is already in a compressed potion with initial compression = 5cm. The 10kg block is **

**allowed to fall. Determine the maximum compression of the spring before the block rebounds. **
**(g = 10ms-2 _{).}**

** **

**29.** **A long plank of mass M = 8kg, length **=1*m*** rests on a horizontal surface. Coefficient of **
**friction **µ1 =0.2**. A small block, mass m = 2kg rests on the right extreme and of M, on the **
**rough top surface. At t = 0, a force F= 25.5N is applied on M towards the right. (see figure). The **
**block m does not slide on M. At t = 3s, force F is increased to 31N. The block m falls off M’s **
**surface at t =7s. Determine the coefficient of friction between m and M.**

** **

**30.** **A particle is released on the smooth inside wall of a cylindrical tank at A with a velocity u which **
**makes an angle **

### α

**with the horizontal tangent. When the particle reaches a point B, a distance**

**h below A, determine the angle**β

**(as an inverse function of cosine) made by its velocity with**

**the horizontal tangent at B.**

** **
**Passage-I (31 to 33):**

**Two blocks of masses 10kg and 5kg are placed on a rough horizontal floor as shoen in figure. **
**The strings and pulley are light and pulley is frectionless. The coefficient of friction between **
**10kg block and surface is 0.3 while that between 5kg block and surface is 0.2. A time varying **
**horizontal force. P=5t Newton (t is in sec) in applied on 5kg block as known. [Take g=10ms-2 _{]}**

** **

**31.** **The motion of block starts at t = t0, then t0 is **

**a) 14s** **b) 8s** **c) 9s** **d) 12s**

**32.** **The friction force between 10kg block and surface at **

2

0
*t*

*t* = **is in between**

**a) zero and 10N** **b) 10N and 35N** **c) 12.5N and 17.5N** **d) 12.5N certain value**

**33.** **The acceleration of 5kg block at t = 2t0 is**

**a) 12 ms-2** ** _{b) }** 2
5
14

_{−}

*ms*

**c)**2 9 14

_{−}

*ms*

**d) 2 ms-2**

**Passage-II (34 to 36):**

**A block of mass 4kg is pressed against a rough wall by two perpendicular horizontal force F1 **
**and F2 as coefficient of static friction between the block and floor is 0.6 and that of kinetic **
**friction is 0.5. [Take g=10ms-2 _{]}**

** **

** **

**34.** **For F1 = 300N and F2 = 100N, find the direction and magnitude of friction force acting on the **
**block.**

**c) 107.7N, making an angle of **
−
5
2
tan 1

**with the horizontal in upward direction.**

**d) 91.6N, making an angle of **
−
5
2
tan 1

**with the horizontal in upward direction.**

**35.** **For F1=150N and F2=100N, find the direction and magnitude of friction force action on block.**

**a) 90N, making an angle of **
−
5
2
tan 1

** with the horizontal in upwards direction **

**b) 75N, making an angle of **
−
5
2
tan 1

** with the horizontal in upwards direction**

**c) 170.7N, making an angle of **
−
5
2
tan 1

** with the horizontal in upwards direction**
**d) Zero**

**36.** **For data of Question No.35, find the magnitude of acceleration of block.**

**a) Zero** **b) 22.5 ms-2** _{c) 26.925 ms}-2_{d) 8.175 ms}-2

**37.** **System shown in figure is in equilibrium and at rest. The spring and string are massless, now **
**the string is cut. The acceleration of mass 2m and m just after the string is cut will be **

** **
**a) **

2
*g*

**upwards, g downwards** **b) g upwards, **

2
*g*

** downwards**

**c) g upwards, 2g downwards** **c) 2g upwards, g downwards**

**1)** **c** **2)** **d** **3)** **c** **4)** **a** **5)** **d**
**6)** **b** **7)** **c** **8)** **c** **9)** **b** **10)** **b**
**11)** **c** **12)** **d** **13)** **a** **14)** **d** **15)** **c**
**16)** **d** **17)** **d** **18)** **c** **19)** **d** **20)** **c**
**21)** **b** **22)** **d** **23)** **c** **24)** **a** **25)** **d**
**26)**
*m*
5
24 **27)** _{β} _{2}_{+}*t*6*N* **28)**

### (

17 1### )

*cm*2 5

_{−}

**29)**µ

_{2}=0.1

**30)** + − 2 1 2 1 cos cos

*u*

*gh*

*u*α

**31)**

**a**

**32)**

**c**

**33)**

**c**

**34)**

**c**

**35)**

**b**

**36)**

**d**

**37)**

**a**

**Solutions **

**1.**i)

*m*11 =

*m*22

**……….(1)**

**ii)**

*m*11+

*M*2 =

*m*22 +0.92 ⇒

*M*1 =0.92

**…………(2)**

**iii)**

*m*11+1.61 =

*m*22+

*M*2

**⇒1.61 =**

*M*2

**………….(3**

**iv) **
** **
2
1
2
1
2
2
2
1
1
1
+ =*m* +*x* ⇒ =*x* ⇒ *x*=
*m*
** From (2) ***M*1=0 .9 2
** ** 1.2 0.9 _{1}0_{.}._{2}9 _{4}3 0.75
2
1
2
1 = ⇒ = = =
⇒
*x* =0.75*kg*

**2.** **(d) Self explanatory [acc = 0, u = 0 **

### ∴

**at rest]**

**3.** **(c) Self explanatory [acc = 0, u = 0 **

### ∴

**at rest]**

**4.** **Uphill: F = ***mg*sinθ+µ1*mg* cosθ
**F.ds = ***mg* *ds*sinθ +µ1 *mg* *ds*cosθ

### ∫

*F*.

*ds*=

*mg*

### ∫

*dy*+µ1

*mg*

### ∫

*dx*1 1 .

*h*

*mg*

*mg*+µ =

**………(1)**

**Downhill:**θ µ θ cos sin 2

*mg*

*mg*

*F*= − −

### ∫

=− + ⇒ = . 2 2 .*ds*

*Fds*

*F*

*ds*

*mgh*

*mg*

*F*µ

_{……….(2)}### ( ) ( )

1 + 2 =### (

µ11 +µ22### )

= ∴*W*

*mg*

**5.**

**Case (i)**

*gR*

*u*

*mg*

*R*

*mu*

*T*

*C*

*A*≥ ⇒ = ⇒ > 2 1 2 45 cos 0

**…………..(1)**

**Case (ii)**

** **
*gR*
*v*
*mg*
*R*
*mv*
*T* *C* _{C}*C* ≥ ⇒ ≥ ⇒ >
2
2
0
−
+
=
2
2
1
1
2
1
2
1_{mu}* _{mv}*2

_{mgR}*C*

*gR*

*gR*

*gR*

*u*

*gR*

*gR*

*u*

*vC*= − + ⇒ − + > ⇒ 2 2 2 2 2 2 2

### (

### )

*gR*

*u*2 3 2 2 = − ⇒

**………….(2)**

### ( )

### ( )

1/ 2 3 2 2### (

3 2 2### )

2 3 1 2 1 2 2 1 2 2_{=}

_{=}−

_{=}

_{−}

_{⇒}

_{=}

_{−}⇒

*u*

*u*

*u*

*u*

**6.** * _{reaction}* = .

*v*.

_{rel}*But*

*v*=0

_{rel}*dt*

*dm*

*F* _{.}

**7.** **No external force horizontally, **

*CM*

*x*

∴ **does not shift**

**Taking **

### →

### (

### )

*m*

*x*

_{A}_{/}

*=0.3sin53−sin37 =0.06 =*

_{B}*C*

*B*

*x*/

**unknown**

*m*

*given*

*xC*= =−0.005

### (

_{/}+

_{/}+

### )

+### (

_{/}+

### )

+### ( )

=0 ∴*mA*

*xA*

*B*

*xB*

*C*

*xC*

*mB*

*xB*

*C*

*xC*

*mC*

*xC*

### (

0.06 0005### ) (

2 0.005### ) (

3 0.005### )

0 1 +_{/}− +

_{/}− + − = ⇒

*xB*

*C*

*xB*

*C*

*C*

*C*

*B*

*B*

*C*

*B*

*x*

*x*

*x*

*x*= ⇒ = + ⇒ / 0.01 / . 005 . 0 005 . 0 01 . 0 − =

*m*=

*cm*

*m*

*B*=0.10+0.05=0.105 =10.5 ⇒

**8.**

**Before collision:**

**Using loss of P.E = Gain of KE**

** **
**Components **
2
3
.
5
−
=
*y*
*u*
2
5
−
=
*x*
*u*
**After collision**

1
4
15
. =+ −
−
= *eu* *ms*
*vy* *y*
1
2
2
4
35
2
5 _{∴} _{=} _{+} _{=} −
−
=
=*u* *v* *v* *v* *ms*
*vx* *x* *y* *x*

**9.** **Work done by friction, wf.**

**We know wf has to be minimum**

### (

### )

*mg*

*mg* µ

µ 0.4+0.3 =0.7,

= **.**

**But wf = loss of P.E = 0.1mg**
7
1
7
.
0
1 ≥ ⇒ <
∴ *mg* µ*mg* µ
**10.**

### ^

### ^

### c o s

*s i n i*

*uj*

*uu*

### θ +

### θ

### =

### ^

### ^

### 1

*e u*

*s i n i*

*uj*

## c o s

*u*

## =

## θ +

## θ

### ^

### ^

### 2

### 2

*ue*

*s i n i*

*uj*

## c o s

*u*

## =

## θ +

## θ

### ^

### ^

### 3

### 3

*ue*

*s i n i*

*uj*

### c o s

*u*

### =

### θ +

### θ

*H*

*e*

*g*

*u*

*e*

*H*

*g*

*e*

*H*6 2 2 6 3 2 2 2 sin 2 sin = = = θ θ

6
2
6
12
36
6
36
12
6
3 10
2
10
2
2
10
1
=
=
⇒
=
=
∴ *e*
*e*
*H*
*H*
64
.
0
100
64
10
2
2
6
=
=
=
⇒*e*
**11.**

### ( )

### ( )

_{2}

*x*

*b*

*a*

*dx*

*x*

*dU*− + − = 0 0

_{2}2 = − − ⇒ ⇒ =

*x*

*b*

*ax*

*um*

*equillibri*

*dx*

*dU*

*a*

*b*

*x*

*and*

*x*≠ =− ⇒

_{0}2

⇒**b and a are of opposite signs. **

⇒** options (b) or (c) possible **
**For stable equation.**

2
2
*dx*
*U*
*d*
** should be +ve**

### ( )( )

− −2._{3}=

_{3}>0⇒

### (

>0### )

>0 ⇒*for*

*x*

*b*

*x*

*b*

*x*

*b*

**12.**

**T’ =70N, T = T’ -30N=40N**

**On cutting thread,**1 1 40 1 40

_{=}− =

*ms*

*A*

_{kg}**All other’s acc = 0**

**13.** *N*_{1} =*N*_{2}sin37+*mg* **……..(1)**

*R*
*m*

*N* 2

5
.
23
1 =
∴*N*
5
.
22
2 =
*N*

### ( )

1 ⇒23.5=22.5×0.6*mg*⇒

*kg*

*m*

*N*

*mg*=23.5−13.5=10 ⇒ =1 ⇒

**14.**

*N*

*fstatic*=20 2 10 20

_{=}= ∴

*A*5 20 25 5 25−

_{=}− = ∴

*a*

*f*⇒ <

*A*

*a*

**not possible**

**Hence no relative motion**

2
66
.
1
5
10
25 _{=} _{−}
+
=
=
⇒*A* *a* *ms*
**15.** **Case(A)**
*a*
*R*
*a*
*R* 40 1 40
1− 1 = ⇒ 1 = −
**Case (B)**
** **
*a*
*R*
*a*
*R* 15 1 15
1− 2 = ⇒ 2 = −
1
2 *R*
*R* >
∴

**16.** **R is clearly true, but A is false. Initial energy is P.E. If wedge is free to move, it will have KE**⇒
**KE of block will be less than that if wedgeis fixed.**

**17.** **You did positive work and gave positive energy. Gravity did equal negative work and made net **

*KE*

∆ ** zero.**

**18.** **Self explanatory.**

**19.** **Tension in rope remains same through out but it is the frictional force provided by the ground **
**that helps a team to win.**

**20.** **Assertion is right because only the charge in kinetic energy in both the frames are same **
**(including work of pseudo force) but kf is dependent on the ki as **

**kf – ki = **∆*k*

**and ki depends on the velocities of these reference frames at time t =0 reason is wrong**
**so assertion is right and reason is wrong**

**21.** **(b) self explanatory **
**22.** **(d) self explanatory**
**23.** **(c) self explanatory**
**24.** **(a) 1st _{ case}**

*uN*

*f*≤ = 37 cos 20

_{(for no motion)}### (

20 20sin37### )

2 . 0 16≤ +*F*+

*N*

*F*≥48

### (

∞### )

∈ 48,*F*

_{………….(1)}**IInd**

_{case}*N*

*f*

*F*

*mg* − sin37 = ≤µ _{(to prevent downward slipping)}

### (

cos37### )

2 . 0 5 3 20−*F*≤

*F*20 5 3 25 4 ≥

_{+}

*F*

*N*

*F*19 500 ≥

*N*
*f*
*mg*

*F*sin37 − = ≤µ _{(to prevent upward slipping)}

≤
−
5
4
2
.
0
20
5
3
*F*
*F*
20
25
4
5
3
≤
−
*F*
∈
≤ *N* *so* *F* *N* *N*
*F*
11
500
,
19
500
11
500
**……..(2)**
**IIIrd _{ case}**

*N*

*f*

*F*

*mg*sin37 − sin37 = ≤µ _{(for no downward slipping)}

_{+}
≤
− 16
5
4
2
.
0
5
3
12 *F* *F*
5
16
12
5
3
25
4 _{≥} _{−}
_{+}
*F*
*N*
*F*
19
220
19
25
5
44_{×} _{=}
≥
*N*
*f*
*F*
*mg* + = ≤µ

− sin37 sin37 _{(for no upward slipping)}

_{+}
≤
− 16
5
4
2
.
0
12
5
3*F* _{F}*N*
*F*
*F*
11
380
11
25
5
76
5
76
25
4
5
3
=
×
≤
⇒
≤
−
**So, **
∈
11
380
,
19
220
*N*
*F* _{………(3)}**IVth _{ Case}**

*N*

*F*

*mg*

*f* = sin37 − cos37 ≤µ _{(for no downward sliding with respect to wedge)}

+
≤
−
5
3
16
2
.
0
5
4
12 *F* *F*
*N*
*F*
*F*
23
220
5
16
12
25
3
5
4 _{≥} _{−} _{⇒} _{≥}
+
*N*
*f*
*mg*

*F*cos37− sin37 = ≤µ _{(for no upward sliding with respect to wedge)}

+
≤
−
5
3
16
2
.
0
12
5
4 *F*
*F*

17
380
≤
*F*
∈ *N* *N*
*F*
17
380
,
19
220
**……….(4)**
**25.** **(d) Self explanatory**

**26.** **When m leaves C, let its velocity (horizontal) be u.**
**Then block 5m will have velocity **

5
*u*
** to left.**

### ∴

**Energy equation:**1 3 50 3 5

_{=}− = ⇒

*u*

*gR*

*ms*

∴_{t = time to reach ground }*s*

*g*
*h*
5
1
10
1
2
2 _{=} × _{=}
=

∴**Distance of hitting point from D =**

*m*
*t*
*u*
*t*
*u*
*t*
*u*
5
24
5
1
.
3
50
5
6
5
6
5× + × = = =
=
**27.** 2 2 4
2
1
2
1
*t*
*v*
*mv*
*K* = = = **………..(1)**
4
2
2
2
2
1
.
1
*t*
*v*
*v*
*R*
*mv*
*F _{n}* = = = =
∴
4
2

_{2t}*v*=

_{v}_{=}

*2*

_{2 t}

_{t}*dt*

*dv*2 2 =

*t*

*dt*

*dv*

*dt*

*dv*

*m*

*F*= =1. =2 2 ∴

_{t}**Resultant**

_{F}*2*

_{F}*2*

_{F}_{2}

_{t}_{2}

*6*

_{t}*n*

*t*+ = + =

**28.**

**Initial energy:**

### (

### )

*J*

*kx*

*E*

*P*

*4000 0.05 5 2 1 2 1 .*

_{spring}_{=}2

_{=}

_{×}

_{×}2

_{=}

*J*

*mgh*

*E*

*P*.

*mass*= =100×0.2=20 ∴

**Total**

*Einitial*=25

*J*

**Final energy**

### (

2### )

_{4000}

### (

_{0}

_{.}

_{05}

_{'}

### )

2 2 1 2 1 .*E*

*k*

*x*

*x*

*x*

*P*

*= + = × × +*

_{spring}### (

0.0025 ' 0.1 '### )

5 2000 ' 200 ' 4000 2 1_{×}

_{×}

_{+}

*2*

_{x}_{+}

_{x}_{+}

*2*

_{x}_{+}

*' 100 ' .*

_{x}*E*

*mgx*

*x*

*P*

*mass*=− =− ∴

_{2000}

_{x}_{'}2

_{+}

_{100}

_{x}_{−}

_{20}

_{=}

_{0}

### (

### )

*m*

### (

### )

*cm*

*x*17 1 2 5 1 17 200 5 '= − = − ⇒

**29.** **From t = 0 to t = 3sec**

**f2 < f2 max**

**so both blocks are moving together and then **

**F – f1 max = (m + M) a** **………(1)**
**f1 max =0.2 x 10 x 10 = 20N**
**So 25.5 – 20 = 10a**
1
/
55
.
0 −
=
⇒*a* *m* *s* **……….(2)**
**Displacement ** 2
1 _{2} 0.55 3
1
×
×
=
*S*
** =2.475m** **……….(3)**
**Velocity = 0.55 x 3 **
** = 1.65m/sec ……….(4)**
**From t = 3 to t = 7 sec**
**f2 =f2 max = 20**µ2
*m*
*a*
8
20
20
31− − µ2 =
*m*
*a*
2
20µ2 =

### (

### )

_{4}2 2 1 1=

*a*−

_{M}*a*8 1 = − ⇒

_{m}*aM*

*am*

**After solving these equation.**
1
.
0
,
1
,
8
9
2 =
=
= *m* µ
*M* *a*
*a*

**30.** **Initial velocity u can be split into tangential component **

*and*

*u*cosα _{vertical component u sin}

### α

∴_{vz at B can be attained by,}

*gh*
*u*
*vz* sin 2
2
2
2 _{=} _{α} _{+}
**Normal reaction **
*r*
*mu*
*N* α
2
2_{cos}
=

∴**tangential component at ***B*=*u*cosα

∴_{velocity at }*B*_{=} *v _{z}*2

_{+}

*u*2cos2

_{α}

_{=}

*u*2

_{+}2

*gh*∴ ∠β

**2 1 2 1 cos cos**

_{is given by }*u*

*gh*

*u*+ − α

**31 to 33.**

**Limitng friction force between 5kg blocks and surface is, ** *fL*_{1} =0.2×5×10 =10*N*

**Limiting friction force between 10kg block and surface is, ** *fL*2 =0.3×10×10=30*N*

**For 5kg block, P-2T -** *f*1 =0**[in equilibrium, i.e., when blocks are not moving]**
**For no motion of blocks, ** *f*1> *fL*1 *and* *f*2 ≤ *fL*2

**So, ** *P*−2*T* ≤ *fL*1 *and* *T* ≤ *fL*2

⇒ *P*≤ *fL*1 +2*fL*2

**So, for motion to take place, ***P*≥ *fL*1 +2*fL*2

⇒ 5 2 10 2 30
2
1
0 = *fL* + *fL* = + ×
*t*
⇒ *t*0 =14*s*
**At ***t* *t* 7*s*
2
0 _{=}

= **, the equation for 5kg and 10kg blocks are **

0 0

2

35− *T*− *f*_{1} = *and* *T*− *f*_{2} =

⇒ 35= *f*1 +2*f*2

**And we know at t = 7s both the blocks are at rest so **

2

1 2

1 *fL* *and* *f* *fL*

*f* ≤ ≤

**Solving above equation we get, **0≤ *f*_{1} ≤10*N*;
*N*
*f*
*N* 17.5
5
.
12 ≤ _{2} ≤
**And ** 12.5≤*T* ≤17.5*N*

**At t = 2t0 = 28s, equation for blocks are **
**140 – 2T – 10 = 5a and T – 30 = 10 x 2a**
⇒ 2
9
14 _{−}
= *ms*
*a*

**34.** **The forces acting on the block are F1, F2, mg, normal contact force and frictional force. Here **
**fractional force won’t act along vertical direction as the component of resultant force along the **
**surface acting on body is not along vertical direction and direction of the friction force is either **
**opposite to the motion of block (direction of acceleration of block) if it is moving or not moving.**

**Resultant of 4g and F2 is 107.7N making an angle of **
−
5
2
tan 1

** with the horizontal **

**As force applied along the surface is **> *fL***, so the block doesn’t move and friction is static in **

**nature.**
*N*
*f* =107.7 **making an angle of **
−
5
2
tan 1

** with the horizontal in upward direction.**
**35.** **For ***F*_{1} =150*N*, *f _{L}* =0.6×150 =90

*N*

**As component of resultant force along the surface is 107.7N and is greater than ** *fL* **, so kinetic **

**friction comes into existence, i.e., friction force acquires the value ** *f* =µ*kN*1 =0.5×150 =75*N*.
**Its direction is opposite to component of resultant force along the surface.**

**36.** **Acceleration of block ** _{8}_{.}_{175} 2
4
75
7
.
107 _{−}
=
−
= *ms*

**37.** **Initially under equilibrium of mass m: T = mg**

**Now, the string is cut. Therefore, T = mg force is decreased on mass m upwards and **
**downwards on mass 2m.**

### ∴

*g*

*m*

*mg*

*a*= =

_{m}**(downwards)**

**And**2 2 2

*g*

*m*

*mg*

*a*

*= =*

_{m}**(upwards)**