Free Objective Test 01 Practice Test - 11th and 12th
Question 1
A force F=−K(yi+aj) (where K is a positive constant) acts on a particle moving in the xy-plane. Starting from the origin, the particle is taken along the positive x-axis to the point (a, 0) and then parallel to the y-axis to the point (a, a). The total work done by the force F on the particles is
−2Ka2
2Ka2
−Ka2
Ka2
SOLUTION
Solution : C
While moving from (0,0) to (a,0)
Along positive x-axis, y = 0 ∴→F=−kx^j
i.e. force is in negative y-direction while displacement is in positive x-direction.
∴W1=0
Because force is perpendicular to displacement
Then particle moves from (a,0) to (a,a) along a line parallel to y-axis (x = +a) during this →F=−k(y^i+a^j)
The first component of force, −ky^i will not contribute any work because this components is along negative x-direction (−^i) while displacement is in positive y-direction (a,0) to (a,a).
The second component of force i.e. −ka^j
∴W2=(−ka^j)(a^j)=(−ka)(a)=−ka2
So net work done on the particle W=W1+W2
=0+(−ka2)=−ka2
Question 2
Which of the following is a scalar quantity
Displacement
Electric field
Acceleration
Work
SOLUTION
Solution : D
Work is a dot product of force and displacement and hence a scalar.
Question 3
If the unit of force and length each be increased by four times, then the unit of energy is increased by
16 times
8 times
2 times
4 times
SOLUTION
Solution : A
Work = Force × Displacement
If unit of force and length be increased by four times then the unit of energy will increase by 16 times.
Question 4
A man pushes a wall and fails to displace it. He does
Negative work
Positive but not maximum work
No work at all
Maximum work
SOLUTION
Solution : C
No displacement is there. Hence work is zero.
Question 5
When work is done on a body by an external force, its
Only kinetic energy increases
Only potential energy increases
Both kinetic and potential energies may increase
Sum of kinetic and potential energies remains constant
SOLUTION
Solution : C
Work done will convert into energies - both Kinetic and Potential
Question 6
The work done in pulling up a block of wood weighing 2 kN for a length of 10m on a smooth plane inclined at an angle of 15∘ with the horizontal is
4.36 kJ
5.17 kJ
8.91 kJ
9.82 kJ
SOLUTION
Solution : B
W=mg sinθ×s
=2×103×sin15∘×10
=5.17kJ
Question 7
A force acts on a 30 gm particle in such a way that the position of the particle as a function of time is given by x=3t−4t2+t3, where x is in metres and t is in seconds. The work done during the first 4 seconds is
5.28 J
450 mJ
490 mJ
530 mJ
SOLUTION
Solution : A
v=dxdt=3−8t+3t2
∴v0=3m/s and v4=19m/s
W=12m(v24−v20) (According to work energy theorem)
=12×0.03×(192−32)=5.28 J
Question 8
A body moves a distance of 10 m along a straight line under the action of a force of 5 N. If the work done is 25 joule, the angle which the force makes with the direction of motion of the body is
0∘
30∘
60∘
90∘
SOLUTION
Solution : C
W=Fscos θ
⇒cos θ=WFs=2550=12
⇒θ=cos−1(12)
⇒θ=60∘
Question 9
A body of mass m is moving in a circle of radius r with a constant speed v. The force on the body is mv2r and is directed towards the centre. What is the work done by this force in moving the body over half the circumference of the circle
mv2πr2
Zero
mv2r2
πr2mv2
SOLUTION
Solution : B
Work done by centripetal force is always zero, because force and instantaneous displacement are always perpendicular.
W=→F.→s=Fs cos θ=Fs cos 90∘=0
Question 10
Given |→A+→B|=|→A−→B| Find the angle between →A and →B
00
450
900
600
SOLUTION
Solution : C
A2+B2+2ABcosθ=A2+B2−2Acosθ⇒4ABcosθ=0⇒cosθ=0⇒θ=90∘
Question 11
A moving body of mass m and velocity 3 km/h collides with a body at rest of mass 2m and sticks to it. Now the combined mass starts to move. What will be the combined velocity
3 km/h
2 km/h
1 km/h
4 km/h
SOLUTION
Solution : C
Initial momentum = m×3+2m×0=3m
Final momentum = 3m×V
By the law of conservation of momentum
3m=3m×V⇒V=1km/h
Question 12
The kinetic energy acquired by a mass m in travelling a certain distance d, starting from rest, under the action of a force F such that the force F is directly proportional to t is
Directly proportional to t2
Independent of t
Directly proportional to t4
Directly proportional to t
SOLUTION
Solution : C
f = kt
m∫v0 dv = k∫t0 tdt
mv = kt2
v = kmt2
KE = 12mv2=12mk2m2t4 = (k22m )t4
KE α t4
Question 13
The blocks A and B shown in fig. Have masses MA=5kg and MB=4kg. The system is released from rest. The speed of B after A has travelled a distance 1 m along the incline is
√32√g
√34√g
√g2√3
√g2
SOLUTION
Solution : C
If A moves down the incline by 1 m, B shal move up by 12m.
If the speed of B is v, then the speed of A will be 2v.
From conservation of energy,
Gain in KE = loss in PE
12mA(2v)2+12mBv2=mAg×35−mBg×12
Solving, we get
v=√g2√3
Question 14
The velocity - time graph of a particle moving in a straight line is shown in figure. The mass of the particle is 2kg.Work done by all the forces acting on the particle in time interval between t = 0 to t = 10 s is
300 J
−300 J
400 J
−400 J
SOLUTION
Solution : A
From work - energy theorem,
W = △KE=Kf−Ki=12m(vf2−vi2)
=12×2[(−20)2−(10)2]=300J
Question 15
The kinetic energy acquired by a mass m in travelling a certain distance d starting from rest under the action of a constant force is directly proportional to
√m
Independent of m
1√m
m
SOLUTION
Solution : B
Kinetic energy acquired by the body
= Force applied on it × Distance covered by the body
K.E=F×d
If F and d both are same then K.E. acquired by the body will be same