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# Laws of Motion (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

## DC Pandey (Questions & Solutions) of Physics: NEET

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## NEET : Laws of Motion (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

The document Laws of Motion (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev is a part of the NEET Course DC Pandey (Questions & Solutions) of Physics: NEET.
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Introductory Exercise 5.1

Ques 1: The diagram shows a rough plank resting on a cylinder with one end of the plank on rough ground. Neglect friction between plank and cylinder. Draw diagrams to show:
(a) the forces acting on the plank,
(b) the forces acting on the cylinder. Sol:
N = Normal force on cylinder by plank R = Normal force on cylinder by ground,
f = Force of friction by ground by cylinder,
w = Weight of cylinder.
N cosθ = f
w + N sinθ = R,
N(R) = Reaction to N, (Force acting on plank)
i.e.,normal force on plank by cylinder
R' = Normal force on plank by ground,
w = Weight of plank,
f' = frictional force on plank by ground.
Resultant of f' and R', N(R) and w pass through point O.

Ques 2: Two spheres A and B are placed between two vertical walls as shown in figure. Friction is absent everywhere. Draw the free body diagrams of both the spheres. Sol: R = Normal force on sphere A by left wall,
N = Normal force on sphere A by ground,
N' = Normal force on sphere A by sphere B,
wA = Weight of sphere A.
N' cos θ = R
N' sin θ + wA = N R' = Normal force on sphere B by right wall,
N ( R) = Reaction to N i.e. normal force on sphere B by sphere A,
wB = Weight of sphere B,
R', N(R) and wB pass through point O, the centre sphere B.

Ques 3: A point A on a sphere of weight W rests in contact with a smooth vertical wall and is supported by a string joining a point B on the sphere to a point C on the wall. Draw free body diagram of the sphere. Sol:
N = Normal force on sphere by wall, w = Weight of sphere,
T = Tension in string.

Ques 4: Write down the components of four forces along ox and oy directions as shown in Fig. 5.33.  Ans:
F1x = 2√3 N, F2x = -2 N, F3x = 0, F4x = 4 N, F1y = 2N, F2y = 2√3 N, F3y = -6 N, F4y = 0
Sol: Component of along x-axis : 4 cos 30° = 2√3 N
along y-axis : 4 sin 30° = 2 N
Component of along x-axis : 4 cos 120° = - 2 N
along y-axis : 4 sin 120° = 2√3 N
Component of along x-axis : 6 cos 270° = 0 N
along y-axis : 6 sin 270° = - 6 N
Component of along x-axis : 4 cos 0° = 4 N
along y-axis : 4 sin 0° = 0 N

Ques 5: A uniform rod AB of weight W is hinged to a fixed point at A It is held in the horizontal position by a string, one end of which is attached to B as shown in Fig. 5.34. Find in terms of W, the tension in the string.
Ans:
W
Sol: Taking moment about point A AB = 1
(T sin 30) 1 = w 1/2
⇒ T = w

Ques 6: In Question 3 of the same exercise the radius of the sphere is a. The length of the string is also a. Find tension in the string.
Ans: Sol: See figure (answer to question no. 3)  T cos 30° = w
or or Ques 7: Find the values of the unknown forces if the given set of forces shown in Fig. 5.35 are in equilibrium.  Ans:
F = 10.16 newton, A/= 2.4 newton
Sol: R cos 30° + 3 = f cos 60° i.e., or
R√3 + 6 = f …(i)
and R sin 30° + f sin 60° = 10
i.e., or R + f√3 = 20 …(ii)
Substituting the value of f from Eq. (i) in Eq. (ii)
R + ( R√3 + 6)√3 = 20
4 R + 6√3 = 20 ∴   f = (2.4) √3 + 6
= 10.16 N

Ques 8: Two beads of equal masses m are attached by a string of length √2a and are free to move in a smooth circular ring lying in a vertical plane as shown in Fig. 5.36. Here, a is the radius of the ring. Find the tension and acceleration of B just after the beads are released to move.
Ans: Sol: At point B (instantaneous vertical acceleration only) ∴  mg - T sin 45° = ma …(i)
At point A (instantaneous horizontal acceleration only)
∴  T cos 45° = ma …(ii)
Combining Eqs. (i) and (ii)
mg - ma = ma Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

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