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Physics NEET

What is magnetic susceptibility and how is it different for diamagnetic, paramagnetic, and ferromagnetic materials?

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Pallavi Pathak

Contributor-Level 10

Magnetic susceptibility (? ) measures in an external magnetic field, how much a material gets magnetized.Paramagnetic materials:? is positive but small.
Diamagnetic materials:? is negative and very small.
Ferromagnetic materials:? is positive and large, often many orders of magnitude greater than 1.

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a year ago

Physics NEET

Why does a bar magnet suspended freely always point in the north-south direction?

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Pallavi Pathak

Contributor-Level 10

Any bar magnet which is freely suspended aligns itself with the Earth's magnetic field. Earth attract the north pole of the suspended magnet because earth behaves like a giant bar magnet with a magnetic south pole near the geographic North Pole.

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a year ago

Physics NEET

What is the difference between magnetic intensity (H) and magnetic field (B)?

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Pallavi Pathak

Contributor-Level 10

The magnetizing force applied to a material is called the Magnetic intensity (H) and the total magnetic effect inside a material is called the magnetic field (B), it includes both the applied magnetic field and the material's response (magnetization).
Following is the formula showing the relation:
B =? (H + M)
Where? is the permeability of free space and M is magnetization.

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Physics Class 12th

A particle of mass m with an initial velocity collides with a particle of mass 3m at rest. After collision, the two particles stick together and the combined particle moves with a velocity . Which of the following is incorrect?

(a) v = v’/2 (b) v = 2v’ (c) v =u/4 (

...more

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New answer posted

a year ago

Physics Class 12th

If mass M, area A' and velocity V are chosen as fundamental units, then the dimension of coefficient of viscosity will be:

(a) M V/ A' (b) MA'/ V (c) MV A' (d) M/ VA'

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Vishal Baghel

Contributor-Level 10

1. $\frac{d q}{d t} = t = 20 t + 8 t^{2}$

$\Rightarrow \int_{0}^{q} d q = \int_{0}^{15} (20 t + 8 t^{2}) d t$

$\Rightarrow q = 20 * \frac{1 5^{2}}{2} + 8 . \frac{1 5^{3}}{3} = 11250 C$

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a year ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Six

Two blocks M₁ and M₂ having equal mass are free to move on a horizontal frictionless surface. M₂ is attached to a massless spring as shown in Fig. 6.10. Initially M₂ is at rest and M₁ is moving toward M₂ with speed v and collides head-on with M₂.

(a) While spring is fully compressed all the KE of M₁ is stored as PE of spring.

(b) While spring is fully compressed the system momentum is not conserved, though final momentum is equal to initial momentum.

(c) If spring is massless, the final state of the M₁ is state of rest.

(d) If the surface on which blocks are moving has friction, then collision cannot be elastic.

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(c) When M1 comes in contact with the spring. M1 is retarded by the spring force and M2 is accelerated by the spring force.

a) So spring will continue compress until the blocks moves with the same velocity.

b) As the surfaces are frictionless momentum of the system will conserved.

c) If the spring is massless whole energy of M1 will be imparted and M1 will be rest .

d) collision is elastic even if friction is not involved.

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New answer posted

a year ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Six

A bullet of mass m fired at 30° to the horizontal leaves the barrel of the gun with a velocity v. The bullet hits a soft target at a height h above the ground while it is moving downward and emerges out with half the kinetic energy it had before hitting the target. Which of the following statements are correct in respect of bullet after it emerges out of the target?

(a) The velocity of the bullet will be reduced to half its initial value.

(b) The velocity of the bullet will be more than half of its earlier velocity.

(c) The bullet will continue to move along the same parabolic path.

(d) The bullet will move in a different parabolic p

...more

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(b) conserving energy between “O” ans ”A”

U_i+ K_i= U_f+ K_f

0+1/2mv²= mgh + 1/2mv'

$\frac{{(v')}^{2}}{2} = \frac{v^{2}}{2} = - g h$

(v')²=v²-2gh = v^'= $\sqrt[]{v^{2} - 2 g h}$ ……….1

Let speed after emerging be v₁then

=1/2mv₁²=1/2[1/2mv'²]

1/2m(v₁)²=1/4m(v')²=1/4m[v²-2gh]

V₁= $\sqrt[]{\frac{v^{2}}{2} - g h}$ ………….2

From eqn 1 and 2

$\frac{v'}{v_{1}} = \frac{\sqrt[]{v^{2} - 2 g h}}{\sqrt[]{v^{2} - 2 g h / \sqrt[]{2}}} = \sqrt[]{2}$

So v₁ = v'/ $\sqrt[]{2}$ =v²(v'/2)

v₁>v'/2

hence after emerging from the target velocity of the bullet is more than half of its earlier velocity v'

(d) as the velocity of the bullet changes to v' which is less than v' hence , path, followed will change and the bullet reaches at point B instead of A

(f) as the bullet is passing through the target

...more

This is a multiple choice answer as classified in NCERT Exemplar

(b) conserving energy between “O” ans ”A”

U_i+ K_i= U_f+ K_f

0+1/2mv²= mgh + 1/2mv'

$\frac{{(v')}^{2}}{2} = \frac{v^{2}}{2} = - g h$

(v')²=v²-2gh = v^'= $\sqrt[]{v^{2} - 2 g h}$ ……….1

Let speed after emerging be v₁then

=1/2mv₁²=1/2[1/2mv'²]

1/2m(v₁)²=1/4m(v')²=1/4m[v²-2gh]

V₁= $\sqrt[]{\frac{v^{2}}{2} - g h}$ ………….2

From eqn 1 and 2

$\frac{v'}{v_{1}} = \frac{\sqrt[]{v^{2} - 2 g h}}{\sqrt[]{v^{2} - 2 g h / \sqrt[]{2}}} = \sqrt[]{2}$

So v₁ = v'/ $\sqrt[]{2}$ =v²(v'/2)

v₁>v'/2

hence after emerging from the target velocity of the bullet is more than half of its earlier velocity v'

(d) as the velocity of the bullet changes to v' which is less than v' hence , path, followed will change and the bullet reaches at point B instead of A

(f) as the bullet is passing through the target the loss in energy of the bullet is transferred to particles of the target . therefore their internal energy increases.

less

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New answer posted

a year ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Six

A man, of mass m, standing at the bottom of the staircase, of height L climbs it and stands at its top.

(a) Work done by all forces on man is equal to the rise in potential energy mgL.

(b) Work done by all forces on man is zero.

(c) Work done by the gravitational force on man is mgL.

(d) The reaction force from a step does not do work because the point of application of the force does not move while the force exists.

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(b, d) When a man of mass m climbs up the staircases of height L, work done by the gravitational force on the man is mgl work done by internal muscular forces will be mgL as the change in kinetic is almost zero.

Hence total work done =-mgL + mgL=0

As the point of application of the contact forces does not move hence work done by reaction forces will be zero.

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New answer posted

a year ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Six

A cricket ball of mass 150 g moving with a speed of 126 km/h hits at the middle of the bat, held firmly at its position by the batsman. The ball moves straight back to the bowler after hitting the bat. Assuming that collision between ball and bat is completely elastic and the two remain in contact for 0.001s, the force that the batsman had to apply to hold the bat firmly at its place would be

(a) 10.5 N

(b) 21 N

(c) 1.05 *104 N

(d) 2.1 * 104 N

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

Time of contact =0.001s

U=126km/h= $\frac{126 * 1000}{60 * 60} = \frac{35 m}{s}$

V= -35m/s

Change in momentum of the ball = m (v-u)= $\frac{3}{20} (- 35 - 35) k g m / s$

=21/2

F= dp/dt=- $\frac{21 / 2}{0.001} N$ = - 1.05 $* 10^{4} N$

Here – negative sign indicates that force will be opposite to the direction of movement of the ball before hitting.

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New answer posted

a year ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Six

Which of the diagrams in Fig. 6.9 correctly shows the change in kinetic energy of an iron sphere falling freely in a lake having sufficient depth to impart it a terminal velocity?

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(b) First velocity of the iron sphere increases and after sometimes becomes constant called terminal velocity. Hence according first KE increases and then becomes constant which is best represented by b.

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