Free Objective Test 02 Practice Test - 11th and 12th
Question 1
For a given velocity, a projectile has the same range R for two angles of projection. If t1 and t2 are the times of flight in the two cases then
SOLUTION
Solution : B
For same range angles of projection should be θ and 90 - θ
So, time of flights t1 = 2usinθg and
t2 = 2usin(90−θ)g = 2ucosθg
By multiplying = t1t2 = 4u2sinθcosθg2
t1t2 = 2g (u2sin2θ)g = 2Rg ⇒ t1t2 α R
Question 2
A body of mass m is thrown upwards at an angle θ with the horizontal with speed v. While rising up the speed of the mass after t seconds will be
SOLUTION
Solution : C
Instantaeous speed of rising mass after t sec will be vt = √vx2+vy2
where vx = v sin θ = Horizontal component of velocity
vy = v sin θ - gt = Vertical component of velocity
vt = √(vcosθ)2+(vsinθ−gt)2
Question 3
A string of length L is fixed at one end and carries a mass M at the other end. The string makes 2π revolutions per second around the vertical axis through the fixed end as shown in the figure, then tension in the string is
SOLUTION
Solution : D
T sin θ = M ω2 R
T sin θ = M ω2 L sin θ
From (i) and (ii)
T = M ω2 L
= M 4π2n2L
= M 4 π2(2π2)L
= 16 ML
Question 4
A stone of mass 1 kg tied to a light inextensible string of length is whirling in a circular path of radius L in a vertical plane. If the ratio of the maximum tension in the string to the minimum tension in the string is 4 and if g is taken to be 10 m/s2 , the speed of the stone at the highest point of the circle is
SOLUTION
Solution : D
Since the maximum tension TB in the string moving in the vertical circle is at the bottom and minimum tension TT is at the top.
TB = mvB2L = mg and TT = mvT2L - mg
∴ TBTT = mvB2L+mgmvT2L−mg or vB2+gLvT2+gL = 41
or vB2 + gL = 4vT2 - 4gL but vB2 = vT2 + 4gL
∴ vT2 + 4gL +gL ⇒ 3 vT2 = 3gL
∴ vT2 = 3 × g × L = 3 × 10 × 103 or vT = 10 m/sec
Question 5
If the sum of two unit vectors is a unit vector, then magnitude of difference is
SOLUTION
Solution : B
Let ^n1 and ^n2 are the two unit vectors, then the sum is
⃗ns = ^n1 + ^n2 or ns2 = n12 + n22 + 2n1n2cosθ
= 1 + 1 + 2cosθ
Since it is given that ns is also a unit vector, therefore 1 = 1+ 1 + 2 cos θ = ⇒ cos θ = - 12 ∴ θ = 120∘
Now the difference vector is ^nd = ^n1 - ^n2 or n12 + n22 - 2n1n2cosθ = 1 + 1 - 2 cos(120∘)
∴ nd2 = 2 - 2(- 12) = 2 + 1 = 3 ⇒ nd = √3
Question 6
The resultant of two vectors A and B is perpendicular to the vector A and its magnitude is equal to half the magnitude of vector B. The angle between A and B is
SOLUTION
Solution : B
B2 = √A2+B2+2ABcosθ ..(i)
∴ tan 90∘ = BsinθA+Bcosθ ⇒ A + Bcosθ = 0
∴ cos θ = - AB
Hence, from (i) B24 = A2+B2−2A2 ⇒ A = √3B2
⇒ cos θ = - AB = - √32 ∴ θ = 150∘
Question 7
A scooter going due east at 10 ms−1 turns right through an angle of 90∘. If the speed of the scooter remains unchanged in taking turn, the change is the velocity of the scooter is
SOLUTION
Solution : D
If the magnitude of vector remains same, only direction change by θ then
⃗△v = ⃗v2 - ⃗v1, ⃗△v = ⃗v2 + (- ⃗v1)
Magnitude of change in vector |⃗△v| = 2 v sin (θ2)
|⃗△v| = 2 × 10 × sin(90∘2) = 10 √2 = 14.14 m/s
Direction is south - west as shown in figure.
Question 8
The maximum and minimum magnitudes of the resultant of two given vectors are 17 units and 7 units respectively. If these two vectors are at right angles to each other, the magnitude of their resultant is
SOLUTION
Solution : D
Rmax = A + B = 17 when θ = 0∘
Rmin = A - B = 7 when θ = 180∘
by solving we get A = 12 and B = 5
Now when θ = 90∘ then R = √A2+B2
⇒R=√(12)2+(5)2 = √169 = 13
Question 9
The sum of the magnitudes of two forces acting at point is 18 and the magnitude of their resultant is 12. If the resultant is at 90∘ with the force of smaller magnitude, what are the, magnitudes of forces
SOLUTION
Solution : C
Let P be the smaller force and Q be the greater force then according to problem-
P + Q = 18 ..(i)
R = √P2+Q2+2PQcosθ = 12 ..(ii)
tan ϕ = QsinθP+Qcosθ = tan 90 = ∞
∴ P + Q cos θ = 0 ..(iii)
By solving (i), (ii) and (iii) we will get P = 5 , and Q = 13
Question 10
A metal sphere is hung by a string fixed to a wall. The sphere is pushed away from the wall by a stick. The forces acting on the sphere are shown in the second diagram. Which of the following statements is wrong
SOLUTION
Solution : B
As the metal sphere is in equilibrium under the effect of three forces therefore →T + →P + →W = 0
From the figure T cos θ =W ...(i)
T sin θ = P ..(ii)
From equation (i) and (ii) we get P= Wtan θ
and T2 = P2 + W2
Question 11
A car is moving on a circular path and takes a turn. If R1 and R2 be the reactions on the inner and outer wheels respectively, then
SOLUTION
Solution : B
Reaction on inner wheel R1 = 12M[g - v2hra]
Reaction on outer wheel R2 = 12M[g + v2hra]
where, r = radius of circular path, 2a = distance between two wheels and h = height of centre of gravity of car
Question 12
The torque of the force →F = (2^i - 3^j + 4^k) N acting at the point →r = (3^i + 2^j + 3^k) m about the origin be
SOLUTION
Solution : B
→τ = →τ × →F= ⎡⎢ ⎢ ⎢⎢^i^j^k3232−34⎤⎥ ⎥ ⎥⎥
= [(2 × 4)-(3 × -3)]^i + (2 × 3)-(3 × 4)]^j + (3 × -3)-(2 × 2)]^k
= 17^i - 6^j - 13^k
Question 13
A man standing on a road holds his umbrella at 30∘ with the vertical to keep the rain away. He throws the umbrella and starts running at 10 km/hr. He finds that raindrops are hitting his head vertically, the speed of raindrops with respect to the road will be
SOLUTION
Solution : B
When the man is at rest w.r.t the the ground, the rain comes to him at an angle 30∘ with the vertical. This is the direction of the velocity
of raindrops with respect to the ground.
Here ⃗vrg = velocity of rain with respect to the ground
⃗vmg = velocity of the main with respect to the ground
and ⃗vrm = velocity of the rain with respect to the man.
We have ⃗vrg = ⃗vrm + ⃗vmg ... (i)
⃗vrg sin 30∘= ⃗vmg = 10 km/hr
or ⃗vrg = 10sin30∘ = 20 km/hr
Question 14
An aeroplane is flying at a constant horizontal velocity of 600 km/hr at an elevation of 6 km towards a point directly above the target on the earth's surface. At an appropriate time, the pilot releases a ball so that it strikes the target at the earth. The ball will appear to be falling
SOLUTION
Solution : C
The pilot will see the ball falling in straight line because the reference frame is moving with the same horizontal velocity but the observer at rest will see the ball falling in parabolic path.
Question 15
The resultant of →P and →Q is perpendicular to →P. What is the angle between →P and →Q.
SOLUTION
Solution : B
⇒ tan90∘ = QsinθP+Qcosθ ⇒ P + Q cos θ = 0
cos θ = −PQ ∴ θ = cos−1(−PQ)
Question 16
A man sitting in a bus travelling in a direction from west to east with a speed of 40 km/h observes that the rain-drops are falling vertically down. To the another man standing on ground the rain will appear
SOLUTION
Solution : B
A man is sitting in a bus and travelling from west to east, and the rain drops are appears falling vertically down.
vm = velocity of man
vr = Actual velocity of rain which is falling at an angle θ with vertical
vrm = velocity of rain w.r.t to moving man
If the another man observe the rain then he will find that actually rain falling with velocity v, at an angle going from west to east.