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10 months ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Eleven

Heat is associated with

(a) Kinetic energy of random motion of molecules

(b) Kinetic energy of orderly motion of molecules

(c) Total kinetic energy of random and orderly motion of molecules

(d) Kinetic energy of random motion in some cases and kinetic energy of orderly motion in other

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(a) We know that temperature increases vibrations of molecules about their mean position increases. Hence kinetic energy associated with random motion of molecules increases.

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Physics Physics NCERT Exemplar Solutions Class 11th Chapter Eleven

As the temperature is increased, the time period of a pendulum

(a) Increases as its effective length increases even though its centre of mass still remains at the centre of the bob

(b) Decreases as its effective length increases even though its centre of mass still remains at the centre of the bob

(c) Increases as its effective length increases due to shifting of centre of mass below the centre of the bob

(d) Decreases as its effective length remains same but the centre of mass shifts above the centre of the bob

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

As the temperature is increased length of the pendulum increases.

T=2 $π \sqrt[]{\frac{L}{g}}$

So if we increase L then T also increases.

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10 months ago

Physics physics ncert exemplar solution class 12th chapter one

Two charges q and -3q are placed fixed on x-axis separated by distance d. where should a third charge 2q be placed such that it will not experience any force?

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alok kumar singh

Contributor-Level 10

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

Answer = $\frac{d}{2}$ (1+ $\sqrt[]{3}$ ) towards left.

Explanation -lets consider 2q charge is placed in between q and -3q, here it will definitely experience some force q charge repel 2q charge and -3q charge will attract 2q charge. So 2q charge will move towards -3q charge.

Now lets consider it to the left of q at some distance x, here the force experience by q is repulsive and force experience by -3q is attractive . so they cancel out each other and no net force is experience by 2q.

Thus, force of attraction by -3q = force of repulsion by q

$?$ k 2q q/x² = k 2q 3q/ (x+d)²

$?$ (x+d)² = 3x

...more

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

Answer = $\frac{d}{2}$ (1+ $\sqrt[]{3}$ ) towards left.

Thus, force of attraction by -3q = force of repulsion by q

$?$ k 2q q/x² = k 2q 3q/ (x+d)²

$?$ (x+d)² = 3x²

$?$ x² + d² + 2xd = 3x²

$?$ 2x² - 2dx – d² = 0

$?$ x = $\frac{d}{2}$ + $\frac{\sqrt[]{3} d}{2}$

$?$ x= $\frac{d}{2}$ (1+ $\sqrt[]{3}$ ) towards left.

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10 months ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Eleven

An aluminium sphere is dipped into water. Which of the following is true?

(a) Buoyancy will be less in water at 0°C than that in water at 4°C.

(b) Buoyancy will be more in water at 0°C than that in water at 4°C.

(c) Buoyancy in water at 0°C will be same as that in water at 4°C.

(d) Buoyancy may be more or less in water at 4°C depending on the radius of the sphere.

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(a) F= V $ρ G$

F $\propto ρ$

$F_{4^{0} C} > F_{0^{0} C}$

Hence buoyancy will be less in water at 0⁰C than that in water at 4⁰C.

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10 months ago

Physics physics ncert exemplar solution class 12th chapter one

Figure represents a crystal unit of cesium chloride, CsCl. The cesium atohis, represented by open circles are situated at the comers of a cube of side 0.40 nm, whereas a Cl atom is situated at
the centre of the Cube. The Cs atoms are deficient in one electron while the Cl atom carries an excess electron.

(i) What is the net electric field on the Cl atom due to eight Cs atoms?
(ii) Suppose that the Cs atom at the comer A is missing. What is the net force now on the Cl atom due to seven remaining Cs atoms?

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alok kumar singh

Contributor-Level 10

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

Explanation- (i)The cesium atoms, are situated at the corners of a cube and Cl atom is situated at the centre of the cube. From the given figure, we can analyse that the chlorine atom is at equal distance from all the eight comers of cube where cesium atoms are placed. Thus, due to symmetry the electric field due to all Cs atoms, on Cl atom will cancel out. Hence net electric field at the centre of cube is zero.

(ii) we know, f=qE E= Kq/r²= Ke/r²

F= e (E)=e (ke/r²)=ke²/r²

Distance= $\sqrt[]{r^{2} + r^{2} + r^{2}}$ = $\sqrt[]{{0.2}^{2} + {0.2}^{2} + {0.2}^{2}} *$ 10^-9m

F=8.99 $*$ 10⁹ (1.6 $*$ 10^-16)²/ (0.346 $*$ 10^-9)²=1.92 $*$ 10^-9N

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10 months ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Eleven

The graph between two temperature scales A and B is shown in Fig. Between upper fixed point and lower fixed point there are 150 equal division on scale A and 100 on scale B. The relationship for conversion between the two scales is given by

(a) $\frac{t_{A} - 180}{100} = \frac{t_{B}}{150}$

(b) $\frac{t_{A} - 30}{150} = \frac{t_{B}}{100}$

(c) $\frac{t_{B} - 180}{150} = \frac{t_{A}}{100}$

(d) $\frac{t_{B} - 40}{100} = \frac{t_{A}}{180}$

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(b) Lowest point for scale A is 30⁰ and lowest point for scale B is 0⁰. Highest point for the scale A is 180⁰ and for scale B is 100⁰

\frac{t_{A} - {L E P}_{A}}{{U F P}_{A} - {L F P}_{A}} = \frac{t_{B} - {L F P}_{B}}{{U F P}_{B} - {L F P}_{B}}

\frac{t_{A} - 30}{180 - 30} = \frac{t_{B} - 0}{100 - 0}

$\frac{t_{A} - 30}{150} = \frac{t_{B}}{100}$ ,LFP= lower fixed point , UFP= upper fixed point $\propto ρ$

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10 months ago

Physics physics ncert exemplar solution class 12th chapter one

Consider a coin of Question 20. ft is electrically neutral and contains equal amounts of positive and negative charge of magnitude 34.8 kC. Suppose that these equal charges were concentrated in two point charges separated by
(i) 1 cm — (1/2 x diagonal of the one paisa coin)
(ii) 100 m (~ length of a long building)
(iii) 10⁶ m (radius of the earth). Find the force on each such point charge in each of the three cases. What do you conclude from these results?

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alok kumar singh

Contributor-Level 10

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

Explanation- (i) F= $\frac{K q 2}{r 2}$ = 9 $*$ 10^{9
$*$} (34.8 $*$ 10³)²/ (10^-2)²=1.09 $*$ 10²³N

(ii) F= $\frac{K q 2}{r 2}$ = 9 $*$ 10^{9
$*$} (34.8 $*$ 10³)²/ (100)²=1.09 $*$ 10¹⁵N

(iii) F= $\frac{K q 2}{r 2}$ = 9 $*$ 10⁹ $*$ (34.8 $*$ 10³)²/ (10⁶)²=1.09 $*$ 10⁷N

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10 months ago

Physics physics ncert exemplar solution class 12th chapter one

A paisa coin is made up of Al-Mg alloy and weight 0.75 g. It has a square shape and its diagonal measures 17 mm. It is electrically neutral and contains equal amounts of positive and negative charges.

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alok kumar singh

Contributor-Level 10

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

Explanation- 1 molar mass of Al has N_A (Avogadro number)= 6.023 $*$ 10²³

M= $\frac{N A}{M} *$ m= 6.023 $*$ 10²³/27 $*$ 0.75 = 1.6 $*$ 10²²

Magnitude of positive and negative charges in one paisa coin = Nze

As aluminium has 13 electron and 13 protons so

= 1.6 $*$ 10²² $*$ 13 $*$ 1.6 $*$ 10^-19=33.28 $*$ 10³C.

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10 months ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Eleven

A uniform metallic rod rotates about its perpendicular bisector with constant angular speed. If it is heated uniformly to raise its temperature slightly

(a) Its speed of rotation increases

(b) Its speed of rotation decreases

(c) Its speed of rotation remains same

(d) Its speed increases because its moment of inertia increases

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

L = angular momentum = Iw = constant

I₁w₁=I₂w₂

Due to expansion of the rod I₂>I₁

$\frac{w_{2}}{w_{1}} = \frac{I_{1}}{I_{2}} 1$

w₂1 so angular velocity decreases.

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10 months ago

Physics physics ncert exemplar solution class 12th chapter one

What will be the total flux through the faces of the cube (Fig. 1.9) with a side of length an if a charge q is placed at

(a) A: a corner of the cube.
(b) B: mid-point of an edge of the cube.
(c) C: centre of the face of the cube.
(d) D: mid-point of 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 – in these type of question imagine another cube along the charge is kept in 3 dimensional aspects.

In this part the charge is situated at the corner of cube here we can placed 7 more cube joining the corner of that cube . so charge is being shared by 8 cubes.

Therefore, the total flux through the faces of the cube= q/8 ε₀