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10 months ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Nine

A steel wire of mass µ per unit length with a circular cross section has a radius of 0.1 cm. The wire is of length 10 m when measured lying horizontal, and hangs from a hook on the wall. A mass of 25 kg is hung from the free end of the wire. Assuming the wire to be uniform and lateral strains << longitudinal strains, find the extension in the length of the wire. The density of steel is 7860 kg m^–3 (Young's modules Y=2*10¹¹ Nm^–2). (b) If the yield strength of steel is 2.5*10⁸ Nm^–2, what is the maximum weight that can be hung at the lower end of the wire?

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Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

When a small element of length dx is considered at x from the load x=0

(a) letT(x) and T(x+dx) are tensions on the two cross sections a distance dx apart then

T(x+dx)+T(x)=dmg= $μ$ dxg

dT= $μ$ gx+C

at x=0 T(0)=mg

C=mg

T(x)= $μ$ gx+Mg

Let length dx at x increases by dr then

Young's modulus Y= stress/strain

$\frac{\frac{T (x)}{A}}{\frac{d r}{d x}} = Y$

$\frac{d r}{d x} = \frac{T (x)}{Y A}$

r= $\frac{1}{Y A} \int_{0}^{L} (μ g x + M g) d x$

= $\frac{1}{Y A} [\frac{μ g x^{2}}{2} + M g x]$ ₀^L

r= $\frac{1}{2 * 10^{11} * π * 10^{- 6}} [\frac{π * 786 * 10^{- 3} * 10 * 10}{2} + 25 * 10 * 10]$

so r = 4 $* 10^{- 3} m$

(b) tension will be maximum at x=L

T= $μ g L$ +Mg=(m+M)g

The force = (yield strength) area= $250 * π N$

(m+M)g= 250 $π$

Mg $\approx 250 π$

M= 25 $π \approx 75 k g$

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10 months ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Eight

The earth is an approximate sphere. If the interior contained matter which is not of the same density everywhere, then on the surface of the earth, the acceleration due to gravity

(a) Will be directed towards the centre but not the same everywhere.

(b) Will have the same value everywhere but not directed towards the centre.

(c) Will be same everywhere in magnitude directed towards the centre.

(d) Cannot be zero at any point.

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alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer-d

Explanation- We treated that the mass of earth is at centre . in this case a=g=0 at centre but if earth is not uniform then value of g is different at different point.

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10 months ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Nine

Consider a long steel bar under a tensile stress due to forces F acting at the edges along the length of the bar (Fig. 9.5). Consider a plane making an angle θ with the length. What are the tensile and shearing stresses on this plane?

(a) For what angle is the tensile stress a maximum?

(b) For what angle is the shearing stress a maximum?

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Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

Let the cross sectional area A . consider the equilibrium of the plane aa'. A force F must be acting on this plane making an angle of $\frac{π}{2} - θ$ with the normal ON. Resolving F into components along the plane (FP) and normal to the plane.

By resolving into components

We get F_p= Fcos $θ$

And F_N= Fsin $θ$

Let area of the face aa' be A' then

A/A'=sin $θ$ so A=A'sin $θ$

The tensile stress = normal force/area=Fsin $θ / A'$

= $\frac{F s i n θ}{A / s i n θ} = \frac{F}{A}$ sin^{2
$θ$}

Shearing stress = parallel foce/Area

= $\frac{F c o s θ}{A / s i n θ} = \frac{F}{A} s i n θ c o s θ$

$\frac{F}{2 A} (2 s i n θ c o s θ = \frac{F}{2 A} s i n 2 θ$

a) For stress to be maximum , sin^{2
$θ$}=1

So $θ$ = $\frac{π}{2}$

b) Shearing stress to be maximum

sin2 $θ = 1$

So $θ$ = $\frac{π}{4}$

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10 months ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Eight

A mass m is placed at P a distance h along the normal through the centre O of a thin circular ring of mass M and radius r. If the mass is removed further away such that OP becomes 2h, by what factor the force of gravitation will decrease, if h = r ?

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alok kumar singh

Contributor-Level 10

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

Explanation- Gravitational force at h height F = $\frac{G M m h}{{{(r}^{2} + h^{2})}^{3 / 2}}$

When mass is displaced upto distance 2h then

F'= $\frac{G M m 2 h}{{{[r}^{2} + {(2 h)]}^{2}]}^{\frac{3}{2}}}$

= $\frac{2 G M m h}{{(r^{2} + 4 h^{2})}^{\frac{3}{2}}}$

When h = r then F= $\frac{G M m h}{[{r^{2} + {(r)}^{2}]}^{3 / 2}}$

So F = $\frac{G M m}{2 \sqrt[]{2 r^{2}}}$

F'= $\frac{2 G M m r}{{(r^{2} + 4 r^{2})}^{\frac{3}{2}}} = \frac{2 G M m}{5 \sqrt[]{5 r^{2}}}$

$\frac{F^{'}}{F} = \frac{4 \sqrt[]{2}}{5 \sqrt[]{5}}$

F' = $\frac{4 \sqrt[]{2}}{5 \sqrt[]{5}} F$

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10 months ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Eight

An object of mass m is raised from the surface of the earth to a height equal to the radius of the earth, that is, taken from a distance R to 2R from the centre of the earth. What is the gain in its potential energy?

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alok kumar singh

Contributor-Level 10

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

Explanation-potential energy of the objects at the surface of the earth is =- $\frac{G M m}{R}$

PE of the object at a height equal to radius of earth = $\frac{G M m}{2 R}$

Gain in potential energy = $\frac{G M m}{2 R}$ - (- $\frac{G M m}{R}$ )= $\frac{G M m}{2 R}$

= $\frac{g R^{2} m}{2 R} = \frac{1}{2} m g R$ (GM=gR²)

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10 months ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Eight

Shown are several curves (Fig. 8.2). Explain with reason, which ones amongst them can be possible trajectories traced by a projectile (neglect air friction).

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alok kumar singh

Contributor-Level 10

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

Explanation-The trajectory of a particle under gravitational force of the earth will be in conic section with the centre of the earth as a focus. Only c meets this requirements

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10 months ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Eight

Show the nature of the following graph for a satellite orbiting the earth.

(a) KE vs orbital radius R

(b) PE vs orbital radius R

(c) TE vs orbital radius R.

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alok kumar singh

Contributor-Level 10

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

Explanation- Orbital speed of the satellite v_o= $\sqrt[]{\frac{G M}{R}}$

KE of the satellite of mass m, E_k = 1/2mv_o²= Gmm/2R

So kinetic energy is inversely proportional to distance.

b)potential energy of a satellite E_p=-GMm/R

so it is also inversely proportional to R

c)total energy of the satellite E=E_k+E_p= $\frac{G M m}{2 R} - \frac{G M m}{R}$ = - $\frac{G M m}{2 R}$

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10 months ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Eight

Two identical heavy spheres are separated by a distance 10 times their radius. Will an object placed at the midpoint of the line joining their centres be in stable equilibrium or unstable equilibrium? Give reason for your answer.

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alok kumar singh

Contributor-Level 10

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

Explanation- Let M be the mass and R be the radius of sphere . an object of mass m be placed at the mid point P of the line joining their centres

Force F₁=F₂=GMm/ (5R)²

If we displace it through x distance

New force towards sphere A, F₁'= GMm/ (5R-x)²

Sphere B F₂' = GMm/ (5R+x)²

As F₁'>F₂' therefore a resultant force (F₁'-F₂') acts on the objects towards sphere A therefore objects start to move towards A hence equilibrium is unstable.

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10 months ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Eight

Mean solar day is the time interval between two successive noon when sun passes through zenith point (meridian). Sidereal day is the time interval between two successive transit of a distant star through the zenith point (meridian). By drawing appropriate diagram showing earth's spin and orbital motion, show that mean solar day is four minutes longer than the sidereal day. In other words, distant stars would rise 4 minutes early every successive day. (Hint: you may assume circular orbit for the earth).

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alok kumar singh

Contributor-Level 10

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

Explanation-Every day the earth advances in the orbit by approximately 1⁰. Then it will have to rotate by 361⁰ to have the sun at zenith point again. So 361⁰ corresponds to 24h

So 1^o corresponds to 24/361= 0.066h=3.99 min =4min

Hence distant stars would rise 4 min early every successive day.

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10 months ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Eight

What is the angle between the equatorial plane and the orbital plane of (a) Polar satellite? (b) Geostationary satellite

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alok kumar singh

Contributor-Level 10

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

Explanation- Angle between the equatorial plane and orbital plane of a polar satellite is 90⁰ and angle between equatorial plane and orbital plane of a geostationary satellite is 0⁰.

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