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Physics physics ncert exemplar solution class 12th chapter one

The three charges q/2, q and q/2 are placed at the corners A, B and C of a square of side 'a' as shown in figure. The magnitude of electric field (E) at the corner D of the square, is

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Payal Gupta

Contributor-Level 10

$E = \frac{k q}{2 a^{2}}$

$E' = \frac{k q}{{(\sqrt[]{2} a)}^{2}} = \frac{k q}{2 a^{2}}$

$E_{N e t} = 2 E c o s 45 ° + E'$

$= \frac{k q}{a^{2}} (\frac{1}{\sqrt[]{2}} + \frac{1}{2})$

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Physics physics ncert exemplar solution class 12th chapter one

Given below are two statements:

Statement – I : A point change is brought in an electric field. The value of electric field at a point near to the change may increase if the charge is positive.

Statement – I : An electric dipole is placed in a non-uniform electric field. The net electric force on the dipole will not be zero.

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Payal Gupta

Contributor-Level 10

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Physics Thermodynamics

Given below are two statements:

Statement – I : When $μ$ moment of an ideal gas undergoes adiabatic change from state $(P_{1}, V_{1}, T_{1})$ to state $(P_{2}, V_{2}, T_{2})$ then work done is w = $\frac{μ R (T_{2} - T_{1})}{1 - γ},$ where $γ = \frac{C_{p}}{C_{v}} a n d$ R = universal gas constant.

Statement – II : In the above case, when work done on the gas, the temperature of the gas would rise.

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Payal Gupta

Contributor-Level 10

In case of adiabatic process

$W o r k d o n e, W = \frac{η R (T_{2} - T_{1})}{1 - Y}$

$A s, Δ Q = Δ U + W$

for adiabatic process :- $Δ Q = 0$

$\Rightarrow Δ U = - W$

So, when work is done by the gas, temperature decreases and when work is done on the gas, temperature rises.

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6 months ago

Physics Gravitation

Two planets A and B of equal mass are having their period of revolutions T_A and T_B such that T_A = 2T_B. These planets are revolving in the circular orbits or radii r_A and r_B respectively. Which out of the following would be the correct relationship of their orbits?

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Payal Gupta

Contributor-Level 10

According to kepler's third law of time period –

${(\frac{T_{A}}{T_{B}})}^{2} = {(\frac{r_{A}}{r_{B}})}^{3} \Rightarrow \frac{r_{A}}{r_{B}} = 4^{1 / 3}$

$\Rightarrow r_{4}^{3} = 4$ $r$ $B$ $3$

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Physics Physics NCERT Exemplar Solutions Class 12th Chapter Seven

Match List – I with List – II

	List – I		List – II
A.	Moment of inertia of solid sphere of radius R about and tangent.	I.	5/3MR²
B.	Moment of inertia of hollow sphere of radius (R) about any tanget	II.	7/5MR²
C.	Moment of inertia of circular ring of radius (R) about its diameter.	III.	1/4MR²
D.	Moment of inertial of circular disc of radius (R) about any diameter.	IV.	1/2MR²

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Payal Gupta

Contributor-Level 10

Use formula for M.I.

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Physics Motion in a Plane

Motion of a particle in x – y plane is described by a set of following equations x = 4sin $(\frac{π}{2} - ω t)$ m and y = 4sin $(ω t) m .$ The path of a particle will be

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Payal Gupta

Contributor-Level 10

$x = 4 s i n (\frac{π}{2} - ω t)$ - (i)

$y = 4 s i n (ω t)$ - (ii)

From (i) and (ii) cos2 $ω t + s i n^{2} ω t = {(\frac{x}{4})}^{2} + {(\frac{y}{4})}^{2} = 1$

$\Rightarrow x^{2} + y^{2} = {(4)}^{2}$

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Physics Laws of Motion

A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration (a) is varying with time t as a = k² rt², where k is a constant. The power delivered to the particle by the force acting on it is given as

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Payal Gupta

Contributor-Level 10

a = k²rt²

$\Rightarrow \frac{v^{2}}{r} = k^{2} r t^{2}$

$\Rightarrow v = k r t$

$a_{t} = \frac{d v}{d t} = k r$

⇒ tangential force, F_t = ma_t = mkr

$∴ P o w e r d e l i v e r e d, P = F_{t} v = (m k r) (k r t) = m k^{2} r^{2} t$

Note → Power delivered by centripetal force will be zero.

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6 months ago

Physics Mechanical Properties of Fluids

Given below are two statements : One is labeled as Assertion A and the other is labeled as Reason R.

Assertion A : Product of Pressure (P) and time (t) has the same dimension as that of coefficient of viscosity.

Reason R : Coefficient of viscosity = $\frac{F o r c e}{V e l o c i t y g r a d i e n t}$

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Payal Gupta

Contributor-Level 10

Pressure * time $= \frac{F t}{A} = [M L T^{- 2}] [T] [L^{- 2}] = [M L^{- 1} T^{- 1}]$

Coefficient of viscosity, $η = [M L^{- 1} T^{- 1}]$

$F_{v i s c o u s} = - η A \frac{Δ v}{Δ y}$

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Physics Class 11th

4.22 $\hat{i}$ and $\hat{j}$ are unit vectors along x- and y- axis respectively. What is the magnitude and direction of the vectors $\hat{i}$ + $\hat{j}$ , and $\hat{i}$ - $\hat{j}$ ? What are the components of a vector A= 2 $\hat{i}$ + 3 $\hat{j}$ along the directions of $\hat{i}$ + $\hat{j}$ , and $\hat{i}$ - $\hat{j}$ ? [You may use graphical method]

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Vishal Baghel

Contributor-Level 10

4.22

Let us consider a vector $\overset{?}{P}$ . The equation can be written as

Px = Py = 1 $|\overset{?}{P}|$ = $\sqrt (P_{x}^{2} + P_{y}^{2})$ $|\overset{?}{P}|$ = $\sqrt (1^{2} + 1^{2}) |\overset{?}{P}|$ = $\sqrt{2}$ …….(i)

So the magnitude of vector $\overset{?}{i}$ + $\overset{?}{j}$ = $\sqrt{2}$

Let $θ$ be the angle made by vector $\overset{?}{P}$ , with the x axis as given in the above figure

$\tan ? θ$ = $(P x / P y) θ$ = $\tan^{- 1} ? (1 / 1)$ , $θ$ = 45 $°$ with the x axis

Let $\overset{?}{Q}$ = $\overset{?}{i}$ - $\overset{?}{j}$

$\overset{?}{Q_{x}} \overset{?}{i}$ – $\overset{?}{Q_{y}}$ $\overset{?}{j}$ = ( $\overset{?}{i}$ – $\overset{?}{j})$

$\overset{?}{Q_{x}}$ = $\overset{?}{Q_{y}}$ = 1

$|\overset{?}{Q}|$ = $\sqrt{{Q_{x}}^{2} + {Q_{y}}^{2}}$ = $\sqrt 2$

Hence $|\overset{?}{Q}|$ = $\sqrt 2$ . Therefore the magnitude of ( $\overset{?}{i}$ + $\overset{?}{j})$ = $\sqrt 2$

Let $θ$ be the angle made

...more

4.22

Let us consider a vector $\overset{?}{P}$ . The equation can be written as

Px = Py = 1 $|\overset{?}{P}|$ = $\sqrt (P_{x}^{2} + P_{y}^{2})$ $|\overset{?}{P}|$ = $\sqrt (1^{2} + 1^{2}) |\overset{?}{P}|$ = $\sqrt{2}$ …….(i)

So the magnitude of vector $\overset{?}{i}$ + $\overset{?}{j}$ = $\sqrt{2}$

Let $θ$ be the angle made by vector $\overset{?}{P}$ , with the x axis as given in the above figure

$\tan ? θ$ = $(P x / P y) θ$ = $\tan^{- 1} ? (1 / 1)$ , $θ$ = 45 $°$ with the x axis

Let $\overset{?}{Q}$ = $\overset{?}{i}$ - $\overset{?}{j}$

$\overset{?}{Q_{x}} \overset{?}{i}$ – $\overset{?}{Q_{y}}$ $\overset{?}{j}$ = ( $\overset{?}{i}$ – $\overset{?}{j})$

$\overset{?}{Q_{x}}$ = $\overset{?}{Q_{y}}$ = 1

$|\overset{?}{Q}|$ = $\sqrt{{Q_{x}}^{2} + {Q_{y}}^{2}}$ = $\sqrt 2$

Hence $|\overset{?}{Q}|$ = $\sqrt 2$ . Therefore the magnitude of ( $\overset{?}{i}$ + $\overset{?}{j})$ = $\sqrt 2$

Let $θ$ be the angle made by vector $\overset{?}{Q}$ , with the x axis as given in the above figure

$\tan ? θ$ = $(Q x / Q y) θ$ = ${- \tan}^{- 1} ? (1 / 1)$ , $θ$ = - 45 $°$ with the x axis

It is given that $\overset{?}{A}$ = 2 $\overset{?}{i}$ + 3 $\overset{?}{j}$ . We can write Ax $\overset{?}{i}$ + Ay $\overset{?}{j}$ = 2 $\overset{?}{i}$ + 3 $\overset{?}{j}$

Comparing the components on both sides, we get

Ax =2 and Ay = 3

$|\overset{?}{A}|$ = $\sqrt{{(2}^{2}} + 3^{2}$ ) = $\sqrt 13$

Let $\overset{?}{A_{x}}$ make an angle $θ$ with the x axis, then

$\tan ? θ$ = (Ax / Ay), $θ$ = $\tan^{- 1} (? 3 / 2)$ = 56.31 $°$

So the angle between (2 $\overset{?}{i}$ + 3 $\overset{?}{j}$ ) and ( $\overset{?}{i}$ + $\overset{?}{j}$ )

$θ'$ = 56.31 – 45 = 11.31 $°$

Component of vector $\overset{?}{A}$ , along the direction of $\overset{?}{P}$ , making an angle $θ$

= ( A cos $θ'$ ) $\overset{?}{P}$

= ( A cos 11.31) $\frac{(\overset{?}{i} + \overset{?}{j})}{\sqrt{2}}$

= $\sqrt{13}$ $* \frac{0.9806}{\sqrt{2}}$ ( $\overset{?}{i}$ + $\overset{?}{j}$ )

= 2.5 $*$ $\sqrt{2}$

= $\frac{5}{\sqrt{2}}$

Let $θ'$ be the angle between the vectors (2 $\overset{?}{i}$ + 3 $\overset{?}{j}$ ) and ( $\overset{?}{i}$ - $\overset{?}{j})$

$θ " = 45 + 56.31 = 101.31 °$

Component of vector $\overset{?}{A}$ , along the direction of $\overset{?}{Q}$ , making an angle $θ$

= (A cos $θ "$ ) $\frac{(\overset{?}{i} - \overset{?}{j})}{\sqrt{2}}$

= $\sqrt{13}$ cos ( $101.31 °) \frac{(\overset{?}{i} - \overset{?}{j})}{\sqrt{2}}$

= -0.5( $\overset{?}{i}$ - $\overset{?}{j})$

= $- \frac{1}{\sqrt{2}}$

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Physics Physics NCERT Exemplar Solutions Class 12th Chapter Seven

If the potential barrier across a p-n junction is 0.6 V. Then the electric field intensity, in the depletion region having the width of 6 * 10^-⁶m, will be________ * 10⁵ N/C.

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Vishal Baghel

Contributor-Level 10

As, Ed = VB

$∴ E = \frac{V_{B}}{d} = \frac{0.6}{6 * 1 0^{- 6}} = 1 * 1 0^{5} N / C$

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