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a year ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Four

A football is kicked into the air vertically upwards. What is its (a) acceleration, and (b) velocity at the highest point?

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- if the football is kicked into the air vertically upwards. Acceleration of football will always vertically downwards and equal to acceleration due to gravity. But when it reaches the highest point its velocity is zero and acceleration is retarding.

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New answer posted

a year ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Four

A ball is thrown from a roof top at an angle of 45° above the horizontal. It hits the ground a few seconds later. At what point during its motion, does the ball have

(a) Greatest speed.

(b) Smallest speed.

(c) Greatest acceleration? Explain

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- at point B it will gain the same speed u and after that speed increases and will be maximum just before reaching point c.

During journey from O to A speed decreases and will be maximum at point A and acceleration is always constant.

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New answer posted

a year ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Thirteen

A vessel of volume V contains a mixture of 1 mole of Hydrogen and 1 mole of Oxygen (both considered as ideal). Let f₁(v)dv, denote the fraction of molecules with speed between v and (v + dv) with f₂(v)dv, similarly for oxygen. Then

(a) f₁(v) + f₂(v) = f(v) obeys the Maxwell's distribution law

(b) f₁(v), f₂(v)will obey the Maxwell's distribution law separately

(c) Neither f₁(v), nor f₂(v) will obey the Maxwell's distribution law

(d) f₂(v) and f1 (v) will be the same

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(b) For a function f (v), the number of molecules n=f (v) which are having speeds between v and v+dv

So f₁ (v) and f₂ (v) will obey's maxwell distribution law

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New answer posted

a year ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Thirteen

1 mole of H₂ gas is contained in a box of volume V = 1.00 m³ at T = 300K. The gas is heated to a temperature of T = 3000K and the gas gets converted to a gas of hydrogen atoms. The final pressure would be (considering all gases to be ideal)

(a) Same as the pressure initially

(b) 2 times the pressure initially

(c) 10 times the pressure initially

(d) 20 times the pressure initially

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(d) When number of molecules breaks then number of moles would become double.

P $\propto n T$ where n is the number of moles and T is temperature of ideal gas

As gases breaks number of moles becomes twcice of initial so n₂ =2n₁

$\frac{P_{2}}{P_{1}} = \frac{n_{2} T_{2}}{n_{1} T_{1}} = \frac{2 n_{1} (3000)}{n_{1} (300)}$ = 20

p₂=20p₁

so 20 times

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New answer posted

a year ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Four

A particle is projected in air at some angle to the horizontal, moves along parabola as shown in Fig., where x and y indicate horizontal and vertical directions, respectively. Show in the diagram, direction of velocity and acceleration at points A, B and C.

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- vsin $θ$ = vertical component

V_x= horizontal component of velocity =vcos $θ$ = constant

V_y = vertical component of velocity =vsin $θ$

Velocity will always be tangential to the curve in the direction of motion and acceleration is always vertically downward and is equal to g.

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New answer posted

a year ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Four

A cyclist starts from centre O of a circular park of radius 1km and moves along the path OPRQO as shown Fig. 4.3. If he maintains constant speed of 10ms–1, what is his acceleration at point R in magnitude and direction?

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- the cyclist covers OPRQO path.

As we know whenever an object performing circular motion, acceleration is called centripetal acceleration and is always directed towards the centre.

so there will be centripetal acceleration a= v²/r

So a= 100/1km= 100/1000=0.1m/s² along RO.

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New answer posted

a year ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Thirteen

Volume versus temperature graphs for a given mass of an ideal gas are shown in Fig. at two different values of constant pressure. What can be inferred about relation between p₁ & p₂?

(a) p₁ > p₂

(b) p₁ = p₂

(c) p₁ < p₂

(d) Data is insufficient.

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(a) Pressure is inversely proportional to slope

So $p_{1}$ $p$ $2 = \frac{m_{2}}{m_{1}} <1$

So p₁ > p₂

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New answer posted

a year ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Four

A man wants to reach from A to the opposite corner of the square C (Fig. 4.10). The sides of the square are 100 m. A central square of 50m * 50m is filled with sand. Outside this square, he can walk at a speed 1 m/s. In the central square, he can walk only at a speed of v m/s (v < 1). What is smallest value of v for which he can reach faster via a straight path through the sand than any path in the square outside the sand?

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alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation – T_sand= $\frac{A P + Q C}{1} + \frac{P Q}{v} = \frac{25 \sqrt[]{2} + 25 \sqrt[]{2}}{1} + \frac{50 \sqrt[]{2}}{v}$

= 50 $\sqrt[]{2} + \frac{50 \sqrt[]{2}}{v} = 50 \sqrt[]{2} (\frac{1}{v} + 1)$

Time taken T_outside= $\frac{A R + R C}{1} s$

AR= $\sqrt[]{75^{2} + 25^{2}} = 25 \sqrt[]{10} m$

RC= AR= $25 \sqrt[]{10} m$

T_outside= 2AR= 50 $\sqrt[]{10} s$

T_sandoutside . so After solving we get velocity is greater v= 0.81m/s

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New answer posted

a year ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Thirteen

A cylinder containing an ideal gas is in vertical position and has a piston of mass M that is able to move up or down without friction (Fig). If the temperature is increased

(a) Both p and V of the gas will change.

(b) Only p will increase according to Charle's law.

(c) V will change but not p.

(d) p will change but not V.

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

P= p_a+Mg/A

A= area of piston

P_a= atmospheric pressure

Mg = weight of piston

When temperature is increases

pV=nRT so volume increases at constant pressure.

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New answer posted

a year ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Thirteen

Boyle's law is applicable for an

(a) Adiabatic process.

(b) Isothermal process.

(c) Isobaric process.

(d) Isochoric process.

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(b) Boyle's law is applicable when temperature is constant

PV=nRT=constant

PV= constant

So pressure is inversely proportional to volume.

So process is called isothermal process.

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