(A) A particle of weight W moves under the action of a force
F = A cos ωt
where A and ω are constant at x = 0 and v = 0 at t = 0 Question 1
The velocity time relation is
$\frac{W}{g}\frac{dv}{dt}=Acos{\omega}t$
$\frac{W}{g}dv=Acos{\omega}tdt$
Integrating it we get
$\frac{W}{g}v+c=\frac{A}{\omega}sin{\omega}t$
at t = 0, v = 0 so c=0
This gives
$v=\frac{Ag}{W\omega}sin{\omega}t$
$\frac{dx}{dt}-\frac{Ag}{W\omega}sin{\omega}t=0$
hence (b) is the correct option
$v=\frac{Ag}{W\omega}sin{\omega}t$
$\frac{dx}{dt}=\frac{Ag}{W\omega}sin{\omega}t $
or,
$dx=\frac{Ag}{W\omega}sin{\omega}tdt$
Integrating and taking x = 0 at t = 0
$x=\frac{Ag}{W\omega^2}(1-cos{\omega}t)$
hence (b) is the correct option
Matrix Match type question
Question 3
A pendulum of length l and mass m is supported from the ceiling of the elevator. Let T_{0} be the time period of oscillation when the elevator is still Column A
(a) elevator is moving up with constant velocity
(b) elevator is moving down with constant velocity
(c) elevator is moving up with constant acceleration
(d) elevator is moving down with constant acceleration Column B
(P) T = T_{0}
(Q) T > T_{0}
(R) T < T_{0}
(S) no appropriate match Solution
Time period of oscillation is still is
$T_0=2\pi\sqrt{\frac{l}{g}}$
$T=2\pi\sqrt{\frac{l}{g\pm a}}$
When elevator moves up or down with constant velocity acceleration of the pendulum a=0 and hence $T=T_0$
When the lift is moving upwards with constant acceleration then the weight W=mg is also moving upwards so force equation is
$F_t-W=ma$
$F_t =ma+W=m(a+g)$
Here in this case apparent weight of the pendulum becomes greater than its true weight due to which T becomes less than $T_0$
Where $F_t$ is the tension in the string and W=mg is the weight of the pendulum
When the lift is moving downwards with constant acceleration then the weight W is also moving upwards so force equation is
$W-F_t=ma$
$F_t =W-ma=m(g-a)$
Where $F_t$ is the tension in the string
Here in this case apparent weight of the pendulum becomes less then its true weight due to which T becomes greater than $T_0$
Hence (a) -> (P);(b) -> (P);(c) -> (R);(d) -> (Q)
Question 4
Let
S_{1} = frame of reference at rest
S_{2} = frame of reference at constant velocity
S_{3} = frame of reference at constant acceleration
S_{4} = frame of reference at uniform circular motion
A block A is at rest as seen from frame of reference S_{1} Column A
(a) S_{1}
(b) S_{2}
(c) S_{3}
(d) S_{4} Column B
(P) ΣF ≠ 0
(Q) ΣF =0
(R) a = 0
(S) a ≠ 0
where ΣF is the resultant force and a is the acceleration of the body Solution
Answers are self explanatory
(a) -> (Q), (R)
(b) -> (Q), (R)
(c) -> (P), (S)
(d) -> (P), (S)
Question 5
Consider the figures given below
Column A
(a) Acceleration of mass m_{2} in (A)
(b) Acceleration of mass m_{1} in (B)
(c) Acceleration of mass m_{2} in (C)
(d) Acceleration of mass m_{2} in (B) Column B
(P) (m_{2}g)/(4m_{1} + m_{2})
(Q) (m_{2}g)/(m_{1} + m_{2})
(R) (2m_{2}g)/(4m_{1} + m_{2})
(S) (m_{2}g)/(2m_{1} + m_{2}) Solution
for system (A)
$m_2g - T = m_2a$
$T = m_1a$
So $m_2g - m_1a = m_2a$
$a = \frac {m_2g}{m_1 + m_2}$
for system (B)
$m_2g - 2T = m_2a$
$T = m_1(2a)$
$m_2g - 4m_1a = m_2a$
$a = \frac {m_2g}{4m_1 +m_2}$
for system (C)
$m_2g - T = m_2a$
$T/2 = 2m_1a$
$T = 4m_1a$
So, $a = \frac {m_2g}{4m_1 +m_2}$
Hence
(a) => (Q) ; (b) => (R) ; (c) => (P) ; (d) => (P)
Paragraph Based questions
(B) A,B, C are the objects as shown above in the figure. A, B, C are 2, 3 and 4 Kg respectively. Coefficient of friction between the blocks are given above in the figure
Let $f_1$ be frictional forces between A & B
$f_2$ be frictional force between B & C
$f_3$ be frictional force between C & surface
Let $a_1$, $a_2$, $a_3$ be the acceleration of A ,B and C respectively Question 6
Which one of the following is true
(a) 0≤ f_{1} ≤ 14, 0≤ f_{2} ≤ 10, 0≤ f_{3} ≤ 9
(b) 0≤ f_{1} ≤ 14, 0≤ f_{2} ≤ 6, 0≤ f_{3} ≤ 4
(c) 0≤ f_{1} ≤ 14, 0≤ f_{2} ≤ 4, 0≤ f_{3} ≤ 6
(d) 0≤ f_{1} ≤ 14, 0≤ f_{2} ≤ 10, 0≤ f_{3} ≤ 6 Solution
Answer is (a)
Maximum Value of $f_1 = \mu m_1g=14$ N
Maximum Value of $f_2 = \mu (m_1+m_2)g=10$ N
Maximum Value of $f_3 = \mu (m_1+m_2+m_3)g=9$ N
Question 7
What is the minimum force F to have so that all part moves with non zero acceleration
(a) 9N
(b) 10N
(c) 14N
(d) 11N Solution
Answer is (a)
For relative motion between A & B
$F > \mu m_1g$
$F > 0.7 \times 2 \times 10 >14N$
For relative motion between B & C
Force on B must
$F > \mu (m_1+m_2)g$
$F> 10N$
For relative motion between C & floor
Force on C must
$F > \mu (m_1+m_2+m_3)g$
F > 9 N
So at F=9N, all objects will start moving with respect to floor
Question 8
If F = 12 N, which of the following is true
(a) $f_1 = 11.2N$, $f_2 = 10N$, $f_3 = 9N$
$a_1= a_2 = a_3 = 0.4 \ m/s^2$
(b) $f_1 = 11.2N$, $f_2= 10N$, $f_3 = 9N$
$a_1 = a_2 = 0.4 \ m/s^2$, $a_3 = 0.25 \ m/s^2$
(c) $f_1 = 11N$, $f_2 = 10N$, $f_3 = 0$
$a_1 = a_2 = a_3= 0.45 \ m/s^2$
(d) none of the above Solution
Answer is (b)
Acceleration will be same for A and B as the F=12N are less then Maximum Frictional force between A and B. So block A and B will move together.
Since Force is greater then the frictional force between B and C, there will be relative motion between them
Let write Force equation for each block separately.
For Block A
$12 - f_1 = 2a$
For Block B
$f_1 - f_2 = 3a$
For Block C
$f2 - 9 = 4a_1$
if $a_1 \ne a$
then $f_2= 10$ N
$10 - 9 = 4a_1$
$a_1 = 0.25 \ m/s^2$
$12 - 10 = 5a$
$2 = 5a$
$a = 0.4 \ m/s^2$
So, $f1 = 11.2$ N
Question 9
what is the minimum force required to have relative motion between B & C object
(a) 11.5
(b) 10
(c) 11.25
(d) none of the above Solution
Ans. (c)
Force equation for Block C
$f_2 - 9 = 4a_1$
for relative motion between B & C
$10 - 9 = 4a_1$
$a = 0.25 \ m/s^2$
Force equation for Block B
$ f_1 - 10 = 3 \times 0.25$
$f_1 = 10.75 \ N$
Force equation for block A
$F - 10.75 = 0.5$
$F = 11.25 \ N$
Other method
At 9 N all block will start moves with same acceleration a
So $F - f_1 = 2a$
$f_1 - f_2 = 3a$
$f_2 - 9 = 4a$
$ \frac {F - 9}{9} = a$
$f_2 = 4a + 9$
$f_2 = 4 [\frac {(F - 9)}{9}] + 9$
Question 10
what is the minimum force required to have relative motion between A & B object
(a) 17.5
(b) 16.6
(c) 14
(d) none of the above Solution
Force equation of Block A
$F - f_1 = 2a$
Force equation of Block B
$f_1 - f_2 = 3a$
Force equation of Block C
$f_2 - 9 = 4a_1$
Now for relative motion between B & C
$10 - 9 = 4a_1$
$a_1 = 0.25 \ m/s^2$
Now for relative motion between A & B, $f_1=14N$, So
$F - 14 = 2a$ for Block A
$14 - 10 = 3a$ for Block B
$F - 14 = 2 \times 4/3$
F = 14 + 2.6
F = 16.6 N
Question 11
for F = 15 N
(a) there will be relative motion between A & B
(b) there will be relative motion between B & C
(c) there will be relative motion between C & surface
(d) none of the above Solution
Correct option are (b) and (c)
It is explained from above two questions
(C)
The pulley is assumed as mass less and friction free. Question 12
Find the acceleration of the block assuming no friction is present
(a) F/2m
(b) F/m
(c) 2F/m
(d) none of the above Solution
Force equation for pulley
2T=F or T=F/2 ----(1)
Force equation for Mass
T = ma ---(2)
Solving equation (1) and (2),we get
a = F/2m
Question 13
If friction force F/4 exist between the block and surface
(a) F/m
(b) F/m
(c) F/4m
(d) none of the above Solution
Force equation for pulley
2T=F or T=F/2 ----(1)
Force equation for Mass
T -F/4 = ma ---(2)
Substituting the value of T from equation (1) in equation (2)
(F/2) - (F/4) = ma
F/4 = ma
a = F/ 4m
(D) A particle of charge Q and mass M with an initial velocity v_{0}i enter an electric field E=-Ej Question 14
What force act in x and y direction
(a) F_{x} = 0, F_{y} = -QE j
(b) F_{x} = 0, F_{y} = QE j
(c) F_{x} = qE i, F_{y} = 0
(d) F_{x} = -qE i, F_{y} = 0 Solution
Force of the charge will be given by F=QE
Now It is given E=-Ej
So , F_{x} = 0, F_{y} = -QE j
Hence (a) is the correct option
Question 15
Velocity at time t is given by
(a) (v_{0} + qE/M)i
(b) v_{0}i + (qEt/M) j
(c) v_{0}i – (qEt/M) j
(d) none of the above Solution
From above question we know that F=-QE j
Or ma=-QE j a=-(QE/m) j
Now velocity equation v=u+at
So v= v0i - (qEt/M) j
Hence (c) is the correct option
Question 16
Let us assume that particle is at origin at t = 0, find the position vector at t = at time t
(a) v_{0}t i – (qEt^{2}/2M) j
(b) v_{0}t i + (qEt^{2}/2M) j
(c) (v_{0}t + qEt^{2}/2M)i
(d) none of the above Solution
$\mathbf{r}=\mathbf{v_0}t+\frac{1}{2}\mathbf{a} t^2$
Substituting the values from above
$\mathbf{r}=v_0t\mathbf{i}-\frac{qEt^2}{2M}\mathbf{j}$
(E)A block of Mass M rests on smooth horizontal surface over which it can move without friction. A body of mass m lies on the block. The coefficient of friction between body and block is K. The force F acts in horizontal direction on the block
Question 17
For what values of F, both the bodies will move together without any relative motion
(a) 0 ≤ F ≤ kMg
(b) 0 ≤ F ≤ kmg
(c) 0 ≤ F ≤ k(M+m)g
(d) None of these Solution
Equation of force for Block
Ma=F-f --(1)
Equation of force for Body
ma=f ---(2)
Eliminating a from equation (1) and (2),we get
f=mF/(m+M)
Now $f \leq kmg$
Or
$F \leq k(M+m)g$
Question 18
When the Force F is sufficient to have relative motion between the block and body, what will the acceleration of block and body?
Equation of force for Block
Mb=F-f --(1)
Equation of force for Body
ma=f ---(2)
When relative motion is there
f=kmg
or
Mb=F-kmg
ma=kmg
or
$a=kg$
$b=\frac{F-kmg}{M}$
Question 19
Find the time in which body will fall from the block when relative motion is present. Assume L is the length of the block
From above question we know,
a=kg
$b=\frac{F-kmg}{M}$
It is obvious b > a
The acceleration of the body relative to block will be directed opposite to the motion and will be equal in magnitude
=(b-a)
$=\frac{F-kmg}{M}-kg$
Now
$L=\frac{1}{2}at^2$
Or $t=\sqrt{\frac{2LM}{F-kg(M+m)}}$
Multiple Choice Questions
Question 20
A spring balance is attached to the roof of the car and a mass m is hanging from it. When the car is standing on the horizontal road, the balance correctly tells us the weight of the mass. The car is travelling with a constant horizontal velocity v on the undulating path defined by the function
where y is the height above road surface.
Find the weight of the mass in spring balance as function of time t. We can assume that while going up on the undulating surface, the car floor remains almost horizontal
Car has uniform horizontal velocity
$\frac{dx}{dt}=v$
So we can say x=vt
Now
$v_y=\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}=\frac{A\pi v}{\lambda}cos{\left(\frac{2\pi x}{\lambda}\right)}$
Now vertical acceleration
$a_y=\frac{dv_y}{dt}=-2A\left(\frac{\pi v}{\lambda}\right)^2sin{\left(\frac{2\pi x}{\lambda}\right)}$
$a_y=-2A\left(\frac{\pi v}{\lambda}\right)^2sin{\left(\frac{2\pi vt}{\lambda}\right)}$
So mass m is in accelerated frame of reference, with respect to this frame, the apparent weight of mass is
$W=m(g+a_y)$
$W=m\left[g-2A\left(\frac{\pi v}{\lambda}\right)^2sin{\left(\frac{2\pi vt}{\lambda}\right)}\right]$
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