All Exams >
Grade 12 >
Mathematics for Grade 12 >
All Questions
All questions of Newton’s Laws of Motion for Grade 12 Exam
A particle P is sliding down a frictionless hemispherical bowl.
It passes the point A at t = 0. At this instant of time, the horizontal component of its velocity is v. A bead Q of the same mass as P is ejected from A at t = 0 along the horizontal string AB, with the speed v. Friction between the bead and the string may be neglected. Let tP and tQ be the respective times taken by P and Q to reach the point B. Then :
- a)tP < tQ
- b)tP = tQ
- c)tP > tQ
- d)
Correct answer is option 'A'. Can you explain this answer?
A particle P is sliding down a frictionless hemispherical bowl.
It passes the point A at t = 0. At this instant of time, the horizontal component of its velocity is v. A bead Q of the same mass as P is ejected from A at t = 0 along the horizontal string AB, with the speed v. Friction between the bead and the string may be neglected. Let tP and tQ be the respective times taken by P and Q to reach the point B. Then :
It passes the point A at t = 0. At this instant of time, the horizontal component of its velocity is v. A bead Q of the same mass as P is ejected from A at t = 0 along the horizontal string AB, with the speed v. Friction between the bead and the string may be neglected. Let tP and tQ be the respective times taken by P and Q to reach the point B. Then :
a)
tP < tQ
b)
tP = tQ
c)
tP > tQ
d)
|
Amrutha Chaudhary answered |
At A the horizontal speeds of both the masses is the same. The velocity of Q remains the same in horizontal as no force is acting in the horizontal direction. But in case of P as shown at any intermediate position, the horizontal velocity first increases (due to N sin θ), reaches a max value at O and then decreases. Thus it always remains greater than v. Therefore tP < tQ.
A wire, which passes through the hole in a small bead, is bent in the form of quarter of a circle. The wire is fixed vertically on ground as shown in the figure. The bead is released from near the top of the wire and it slides along the wire without friction. As the bead moves from A to B, the force it applies on the wire is 
- a)always radially outwards
- b)always radially inwards
- c)radially outwards initially and radially inwards later
- d)radially inwards initially and radially outwards later
Correct answer is option 'D'. Can you explain this answer?
A wire, which passes through the hole in a small bead, is bent in the form of quarter of a circle. The wire is fixed vertically on ground as shown in the figure. The bead is released from near the top of the wire and it slides along the wire without friction. As the bead moves from A to B, the force it applies on the wire is
a)
always radially outwards
b)
always radially inwards
c)
radially outwards initially and radially inwards later
d)
radially inwards initially and radially outwards later
|
Aravind Nambiar answered |
As the bead is moving in the circular path
By energy conservation, 
From (1) and (2)
[N = mg cos θ - m [2g - 2g cosθ]
N = mg cos θ - 2mg + 2mg cosθ
N = 3mg cos θ-2mg
⇒ N = mg (3 cos θ-2)
[N = mg cos θ - m [2g - 2g cosθ]
N = mg cos θ - 2mg + 2mg cosθ
N = 3mg cos θ-2mg
⇒ N = mg (3 cos θ-2)
Clearly N is positive (acts radially outwards) when 
Similarly, N acts radially inwards if 
A small block of mass of 0.1 kg lies on a fixed inclined plane PQ which makes an angle θ with the horizontal. A horizontal force of 1 N acts on the block through its centre of mass as shown in the figure.The block remains stationary if (take g = 10 m/s2)
- a)θ = 45°
- b)θ > 45° and a frictional force acts on the block towards P.
- c)θ > 45° and a frictional force acts on the block towards Q.
- d)θ < 45° and a frictional force acts on the block towards Q.
Correct answer is option 'A,C'. Can you explain this answer?
A small block of mass of 0.1 kg lies on a fixed inclined plane PQ which makes an angle θ with the horizontal. A horizontal force of 1 N acts on the block through its centre of mass as shown in the figure.
The block remains stationary if (take g = 10 m/s2)
a)
θ = 45°
b)
θ > 45° and a frictional force acts on the block towards P.
c)
θ > 45° and a frictional force acts on the block towards Q.
d)
θ < 45° and a frictional force acts on the block towards Q.
|
|
Aravind Nambiar answered |
The forces are resolved as shown in the figure.
When θ = 45º, sinθ = cosθ
When θ = 45º, sinθ = cosθ
The block will remain stationary and the frictional force is zero.
When θ > 45º, sinθ > cosθ
Therefore a frictional force acts towards Q.
A simple pen dulum of length L an d mass (bob) M is oscillating in a plane about a vertical line between angular limit – φ and + φ . For an angular displacement θ (|θ| < φ), the tension in the string and the velocity of the bob are T and V respectively. The following relations hold good under the above conditions :- a)T cos θ = Mg.
- b)
- c)The magnitude of the tangenial acceleration of the bob |aT| = g sin θ
- d)T = Mg cos θ
Correct answer is option 'B,C'. Can you explain this answer?
A simple pen dulum of length L an d mass (bob) M is oscillating in a plane about a vertical line between angular limit – φ and + φ . For an angular displacement θ (|θ| < φ), the tension in the string and the velocity of the bob are T and V respectively. The following relations hold good under the above conditions :
a)
T cos θ = Mg.
b)
c)
The magnitude of the tangenial acceleration of the bob |aT| = g sin θ
d)
T = Mg cos θ
|
Anand Khanna answered |
Since the body is moving in a circular path therefore it needs centripetal force 
Also, the tangential acceleration acting on the mass is g sin θ.
A reference frame attached to the earth- a)is an inertial frame by definition.
- b)can not be an inertial frame because the earth is revolving round the sun.
- c)is an inertial frame because Newton’s laws are applicable in this frame.
- d)cannot be an inertial frame because the earth is rotating about its own axis.
Correct answer is option 'B,D'. Can you explain this answer?
A reference frame attached to the earth
a)
is an inertial frame by definition.
b)
can not be an inertial frame because the earth is revolving round the sun.
c)
is an inertial frame because Newton’s laws are applicable in this frame.
d)
cannot be an inertial frame because the earth is rotating about its own axis.
|
|
Aravind Nambiar answered |
Since earth is an accelerated frame and hence, cannot be an inertial frame.
Note : Strictly speaking Earth is accelerated reference frame. Earth is treated as a reference frame for practical examples and Newton's laws are applicable to it only as a limiting case.
Note : Strictly speaking Earth is accelerated reference frame. Earth is treated as a reference frame for practical examples and Newton's laws are applicable to it only as a limiting case.
Chapter doubts & questions for Newton’s Laws of Motion - Mathematics for Grade 12 2025 is part of Grade 12 exam preparation. The chapters have been prepared according to the Grade 12 exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for Grade 12 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
Chapter doubts & questions of Newton’s Laws of Motion - Mathematics for Grade 12 in English & Hindi are available as part of Grade 12 exam.
Download more important topics, notes, lectures and mock test series for Grade 12 Exam by signing up for free.
Mathematics for Grade 12
174 videos|103 docs|119 tests
|
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up
within 7 days!
within 7 days!
Takes less than 10 seconds to signup