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Test: Solving Problems in Mechanics (NCERT) - NEET MCQ


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Test: Solving Problems in Mechanics (NCERT) - Question 1

An iron block of sides 50 cm x 8 cm x 15 cm has to be pushed along the lloor. The force required will be niinimum_when the surface in contact with ground is

Detailed Solution for Test: Solving Problems in Mechanics (NCERT) - Question 1

Force will be same for all the surfaces.

Test: Solving Problems in Mechanics (NCERT) - Question 2

Figure shows a man of mass 55 kg standing stationary with respect to a horizontal conveyor belt that is accelerating with 1 m s-1. The net force acting on the man is 

Detailed Solution for Test: Solving Problems in Mechanics (NCERT) - Question 2

Here,
Mass of the man, M = 55 kg
As the man is standing stationary w.r.t.  the belt,
Acceleration of man = Acceleration of belt
Acceleration of man a = 1ms−2 
Net force on the man,
F = Ma = (55 kg) × (1 m s-2) = 55N
Force on man is 55N 

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Test: Solving Problems in Mechanics (NCERT) - Question 3

A helicopter of mass 2000 kg rises with a vertical acceleration of 15 m s-2. The total mass of the crew and passengers is 500 kg. Choose the correct statements from the following. (Take g = 10 m s-2)
(i) The force on the floor of the helicopter by the crew and passengers is 1.25 x 104 N vertically downwards.
(ii) The action of the rotor of the helicopter on the surrounding air is 6.25 x 104 vertically downwards.
(iii) Theforceon the helicopter due to the surrounding air is 6.25 x 10N vertically upwards.

Detailed Solution for Test: Solving Problems in Mechanics (NCERT) - Question 3

Here,
Mass of hehcopter, M = 2000 kg
Mass of the crew and passengers, m = 500 kg
Vertically upwards acceleration, a = 15 ms-2
g = 10 m s-2
(i) Force on the floor by the crew and passengers
= m (g + a) = 500 kg (10 + 15) m s-2
= 12500 N = 1.25 x 104 N
It acts vertically downwards.
(ii) Action of the rotor of the hehcopter on the surrounding air = (M + m) (g + a)
= (2000 + 500) kg (10 + 15) m s -2
= 62500 N = 6.25 x 104 N
It acts vertically downwards
(iii) Force on the helicopter due to the surrounding air is equal and opposite to the action of the rotor of the helicopter on the surrounding air.
Force on the helicopter due to the surrounding air
= 6.25 x 104 N
It acts vertically upwards.

Test: Solving Problems in Mechanics (NCERT) - Question 4

A person in an elevator accelerating upwards with an acceleration of 2 m s-2, tosses a coin vertically upwards with a speed of 20 m s-1. After how much time will the coin fall back into his hand? (Take g = 10 m s-2)

Detailed Solution for Test: Solving Problems in Mechanics (NCERT) - Question 4

Here, v = 20m s−1, a = 2m s−2, g = 10m s−2
The coin will fall back into the person’s hand after t s.
∴  t = 
= 10/3s

Test: Solving Problems in Mechanics (NCERT) - Question 5

A person of mass 50 kg stands on a weighing scale on a lift. If the lift is ascending upwards with a uniform acceleration o f 9 m s-2, what would be the reading of the weighing scale? (Take g = 10m s-2)

Detailed Solution for Test: Solving Problems in Mechanics (NCERT) - Question 5

The reading on the scale is a measure of the force on the floor by the person. By the Newton’s third law this is equal and opposite to the normal force N on the person by the floor.
∴ When the lift is descending downward with a acceleration of a 9 ms−2, then
N - 50 × 10 = 50 × 9
or N = 50 × 10 + 50 × 9 = 50(10 + 9)
= 950 N
∴  The reading of weighing machine is 95kg.

Test: Solving Problems in Mechanics (NCERT) - Question 6

Block A of weight 100 N rests on a frictionless inclined plane of slope angle 30° as shown in the figure. A flexible cord attached to A passes over a frictionless pulley and is connected to block B of weight W. Find the weight W for which the system is in equilibrium.

Detailed Solution for Test: Solving Problems in Mechanics (NCERT) - Question 6

As the system is in equilibrium,
∴ T = W …(i)
and T = mg sin30 …(ii)
From (i) and (ii), we get

W = mg sin 30 = (100 N) × 1/2 = 50 N

Test: Solving Problems in Mechanics (NCERT) - Question 7

Two blocks of masses 10 kg and 20 kg are connected by a massless string and are placed on a smooth horizontal surface as shown in the figure. If a force F = 600 N is applied to 10 kg block, then the tension in the string is

Detailed Solution for Test: Solving Problems in Mechanics (NCERT) - Question 7

Here, m(1) = 10kg, m = 20kg, F = 600N
Let T be tension of the string and a be common acceleration of the system.
a = 
= 20 ms-2
When force F is applied on 10kg block, then the tension in the string is
T2 = m2a = (20kg)(20ms−2) = 400N
image

Test: Solving Problems in Mechanics (NCERT) - Question 8

Two blocks of masses 10kg and 20kg are connected by a massless string and are placed on a smooth horizontal surface as shown in the figure. If force F = 600N is applied to 20kg block, then the tension in the string is

Detailed Solution for Test: Solving Problems in Mechanics (NCERT) - Question 8



Given

Mass of body (m2) = 10kg

Mass of body (m1) = 20kg

Force applied =600N

Let a is the acceleration of the system.

When force F applied on A,

using free body diagram

For body m1

F−T = m1a ............(1)

for body m2,

T = m2a ................(2)

solving equation (1) and (2),

a = F (m1+m2)

a=600(10+20)

a=20m/s2

Now, 600−T=20∗20........................(Using eqn. (1))

T=200N

 

Test: Solving Problems in Mechanics (NCERT) - Question 9

Two masses of 5 kg and 3 kg are suspended with the help of massless inextensible strings as shown in figure. The whole system is going upwards with an acceleration of 2 m s-2. The tensions T( and T2 are respectively (Takeg = 10 m s-2)

Detailed Solution for Test: Solving Problems in Mechanics (NCERT) - Question 9

The free body diagram of 3 kg block is as shown in the fig. (a).
The equation of motion of 3 kg block is T2 − 3g = 3a 
T2 = 3(a + g) = 3(2 + 10) = 36N     ............(i)
The free body diagram of 5 kg is as shown in the Fig.(b).

The equation of motion of 5kg block is 
T1 −T2 − 5g = 5a 1
= 5(a + g) + T2
= 5(2 + 10) + 36 = 96N         (Using (i))

Test: Solving Problems in Mechanics (NCERT) - Question 10

Two blocks each of mass M are resting on a frictionless inclined plane as shown in figure. Then

Detailed Solution for Test: Solving Problems in Mechanics (NCERT) - Question 10

The force acting down the plane for blocks A and B are:

Since FA > FB the block A moves down the plane.

Test: Solving Problems in Mechanics (NCERT) - Question 11

In the system shown in the figure, the acceleration of 1 kg mass is

Detailed Solution for Test: Solving Problems in Mechanics (NCERT) - Question 11

If a is downward acceleration of 4 kg block, the upward acceleration of 1 kg block must be 2a.
image
If T is tension in each part of string, then equations of motion of the two blocks are
4a = 4g − 2T ...(i)
1 × 2a = T −1 g ..(ii)
or 4a = 2T −2g ...(iii)
Adding (i) and (iii), we get
8a  = 2g, a = g / 4
∴ Acceleration of 1 kg block = 2a = g/2 upwards.

Test: Solving Problems in Mechanics (NCERT) - Question 12

Two blocks of masses 8 kg and 12 kg are connected at the two ends of a light inextensible string. The string passes over a frictionless pulley. The acceleration of the system is

Detailed Solution for Test: Solving Problems in Mechanics (NCERT) - Question 12

Let a be the common acceleration of the system and T be tension of the string. The equations of motion of two blocks are
T − 8g = 8a …(i)
and 12g −T = 12a
image
Adding (i) and (ii), we get
4g = 20a  or a = g/5

Test: Solving Problems in Mechanics (NCERT) - Question 13

A monkey of mass 40 kg climbs on a massless rope which can stand a maximum tension of 500 N. In which of the following cases will the rope break? (Take g = 10 m s-2)
Physics Question Image

Detailed Solution for Test: Solving Problems in Mechanics (NCERT) - Question 13

Here, mass of monkey, m = 40kg
Maximum tension the rope can stand, T = 500N
Tension in the rope will be equal to apparent weight of the monkey (R).
The rope will break when R exceeds T.(a) When the monkey climbs up with an acceleration a = 5m s−2
R = m (g + a) = 40(10 + 5) = 600N
∴  R > T
Hence, the rope will break.
(b) When the monkey climbs down with an acceleration a=5m s−2
R = m (g − a) = 40(10 − 5) = 200N
∴ R < T
Hence, the rope will not break.
(c) When the monkey climbs up with a uniform speed v = 5m s−1, its acceleration a = 0
∴ R = mg = 40 × 10 = 400N
∴ R < T
Hence, the rope will not break.
(d) When the monkey falls down the rope freely under gravity
a = g
∴ R = m(g − a) = m(g − g) = zero
Hence, the rope will not break.

Test: Solving Problems in Mechanics (NCERT) - Question 14

Two blocks of masses of 40 kg and 30 kg connected by a weightless string passing over a frictionless pulley as shown in the figure.
image
The acceleration of the system would be 

Detailed Solution for Test: Solving Problems in Mechanics (NCERT) - Question 14

Here, m1 = 40kg
image
m2 = 30kg, θ = 30
Let T be the tension in the string and a be the acceleration of the system.
Their equations of motion are m1gsin30 −T = m1a …(i)
T − m2gsin30 = m2a …(ii)
Adding (i) and (ii), we get
(m1 + m2) a = (m1 − m2) gsin30
Substituting the given values, we get
(40 + 30) a = (40 − 30) × 9.8 × 1/2 = 49
a = 49/70 = 0.7m s−2

Test: Solving Problems in Mechanics (NCERT) - Question 15

A book is lying on the tab le. What is the angle  between the action of the book on the table and the reaction of the table on the book?

Detailed Solution for Test: Solving Problems in Mechanics (NCERT) - Question 15

The weight (W) and the action (A) both will act in same direction and parallel to each other. So the angle between them should be zero.

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