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6 months ago

Overview Physics Current Electricity

Resistance of given carbon resistor is

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Vishal Baghel

Contributor-Level 10

$01 \times 1 0^{4} Ω \pm 5 % = 10 \times 1 0^{3} Ω \pm 5 %$

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6 months ago

Physics NCERT Exemplar Solutions Class 11th Chapter Eight Physics

Deviation of light ray by a thin prism of refractive index $μ$ and refracting angle $A$ is

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alok kumar singh

Contributor-Level 10

For thin prism, deviation $= (μ - 1) A$

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6 months ago

Physics Class 11th

If force (F), mass (M), length (L) are taken as fundamental quantities, dimensions of time will be

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Vishal Baghel

Contributor-Level 10

$F = m a = \frac{M L}{T^{2}}$

$\Rightarrow T^{2} = \frac{M L}{F}$

$\Rightarrow [T] = ⌊ M^{1 / 2} L^{1 / 2} F^{- 1 / 2} ⌋$

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6 months ago

Physics NCERT Exemplar Solutions Class 11th Chapter Eight physics ncert solutions class 11th

The radius of circle, the period of revolution $T$ , initial position and sense of revolution are indicated in the figure. x-projection of the radius vector of rotating particle $P$ is

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alok kumar singh

Contributor-Level 10

$ω = \frac{2 π}{T} = \frac{π}{5} r a d / s$

$\bar{θ} (a t t = 0) = 0^{0}$

$x = 5 \sin ? ω t$

$x = 5 \sin ? (\frac{π}{5} t)$

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6 months ago

Physics Ncert Solutions Class 12th Semiconductor Electronics: Materials, Devices and

What is the current through an ideal P – N junction diode shown in figure below?

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Vishal Baghel

Contributor-Level 10

$\Rightarrow l = \frac{2}{50} = 40 m A$

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6 months ago

Engineering VTUEEE

When will VTUEEE registrations start?

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Nishtha Chaudhary

Contributor-Level 7

The VTUEEE registrations is expected to start soon. We will share all the dates of the VTUEEE registrations, as and when it is out.

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6 months ago

Physics Motion in Straight Line Physics

The velocity of a particle is given by $v = \sqrt[]{180 - 16 x} m / s$ . Its acceleration will be

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Raj Pandey

Contributor-Level 9

$V = \sqrt[]{180 - 16 x} m / sec$

$ac c^{n} = V \frac{d v}{d x}$

$= {(180 - 16 x)}^{1 / 2} \frac{d {(180 - 16 x)}^{1 / 2}}{d x}$

= ${(18 - 16 x)}^{1 / 2} \times \frac{1}{2} {(180 - 16 x)}^{1 / 2} \times - 16$

= $? 8 m / s e c^{2}$

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6 months ago

Alternating Current Overview Physics

The value of alternating emf (E) in the given circuit will be

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Vishal Baghel

Contributor-Level 10

$E = \sqrt[]{V_{R}^{2} + {(V_{L} - V_{C})}^{2}}$

$E = \sqrt[]{{(60)}^{2} + {(100 - 20)}^{2}}$

E = 100V

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6 months ago

Physics NCERT Exemplar Solutions Class 11th Chapter Eight Physics

Two conducting rods inclined at $' θ'$ angle from horizontal are kept parallel at separation $' l'$ to each other. A resistor of resistance $R$ joins the two rods at their topmost point. A metal bar of length $' l'$ is free to slide on the rods as shown in figure. on releasing from rest, the maximum speed that can be attained by the bar if a constant magnetic field $B$ exists vertically downwards, is

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alok kumar singh

Contributor-Level 10

$F = i l \times B$

$? m g \sin ? θ = i l \times B$

$? m g \sin ? θ = (\frac{B \cos ? θ l v}{R}) l B \cos ? θ$

$? V = \frac{R m g \sin ? θ}{B^{2} l^{2} \cos^{2} ? θ}$

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6 months ago

Physics Class 12th

A farsighted person cannot focus distinctly objects closer than 100cm. The lens that will permit him to read from a distance of 50cm, will have a focal length:

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Vishal Baghel

Contributor-Level 10

$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$

$\Rightarrow \frac{1}{f} = \frac{1}{- 100} - \frac{1}{- 50} = \frac{1}{50} - \frac{1}{100}$

$\Rightarrow f = + 100 c m$

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6 months ago

Electrostatics Physics Electrostatic Potential and Capacitance

Figure shown four plates each of area A and separated from another by a distance d. What is the capacitance between P and Q?

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Vishal Baghel

Contributor-Level 10

$C_{e q} = 2 C = \frac{2 A ε_{0}}{d}$

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6 months ago

Scalars and Vectors Physics Motion in Plane

If $a \cdot b = | a \times b |$ , then the $θ$ angle between a and b will be

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Raj Pandey

Contributor-Level 9

$a \cdot b = | a \times b | a b$ $cos θ = a b sin θ$

$\frac{\underline{a b}}{\underline{a b}} = \frac{s i n θ}{c o s θ}$

$tan θ = 1$

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6 months ago

Physics NCERT Exemplar Solutions Class 11th Chapter Eight Overview

Consider the following two statements .

(A)The gravitational force on a particle inside a spherical shell is zero.

(B) Total mechanical energy of a body on earth’s surface may be positive.

Which of the following options is correct?

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alok kumar singh

Contributor-Level 10

Inside a spherical shell gravitational field is zero, so net force acting on a particle placed inside it is also zero.

Total energy of a bound system is always negative.

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6 months ago

Physics Class 11th

A body possesses kinetic energy x, moving on a rough horizontal surface, is stopped in a distance y. The friction force exerted on the body is

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Vishal Baghel

Contributor-Level 10

Work done = $Δ K E$

$\Rightarrow - f = - x$

$\Rightarrow f = \frac{x}{y}$

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6 months ago

Physics Motion in Plane Physics NCERT Exemplar Solutions Class 11th Chapter Eight

A man who can swim with a speed of $2 k h / h$ wants to cross a river $200 m$ wide. If the river is flowing with a speed of $3 k m / h$ then in what minimum time can the man cross the river?

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alok kumar singh

Contributor-Level 10

Time (minimum) taken to cross the river

$t = \frac{W}{V_{s / r}} = \frac{200 \times 10^{- 3}}{2}$

$100 \times 10^{- 3} h$

$= 0.1 h = 6 m i n u t e s$

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6 months ago

Electrostatics Physics Electrostatic Potential and Capacitance

A charged particle of mass m and charge q is released from rest in a uniform electric field E. The KE of particle after time t is:

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Vishal Baghel

Contributor-Level 10

$F = q E a = \frac{q E}{m}$

$\Rightarrow s = \frac{1}{2} a t^{2} = \frac{1}{2} (\frac{q E}{m}) t^{2}$

$\Rightarrow W o r k d o n e = Δ K$

$\Rightarrow Δ K = F . s = \frac{q^{2} E^{2} t^{2}}{2 m}$

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6 months ago

Overview Physics

For the arrangement shown in figure, the tension in the string to prevent it from sliding down, is (g = 10 m/s²)

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Vishal Baghel

Contributor-Level 10

$\begin{array}{l} f_{l} = μ N = 0.6 \times m g c o s θ \\ f_{l} = 0.6 \times 2 \times 10 \times \frac{4}{5} = 9.6 N \end{array}$

Component of weight along incline.

$= m g s i n θ = 2 \times 10 \times \frac{3}{5} = 12 N$

$\Rightarrow m g s i n θ > f_{l} s o T = 2.4 N$

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6 months ago

Potential Energy of Spring Physics NCERT Exemplar Solutions Class 11th Chapter Eight

Two springs with spring constants $k_{1} = 1200 N / m$ and $k_{2} = 500 N / m$ are stretched by the same force. The ratio of potential energy in the springs will be

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alok kumar singh

Contributor-Level 10

Elastic potential energy stored in the spring

$U = \frac{1}{2} k x^{2}$

$= \frac{1}{2} F x = \frac{F^{2}}{2 k}$

$H e n c e \frac{U_{1}}{U_{2}} = \frac{k_{2}}{k_{1}} [F i s s a m e f o r b o t h s p r i n g s]$

$= \frac{500}{1200} = \frac{5}{12}$

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6 months ago

Significant Figures Physics

Two rods with lengths 20.123 cm and 18.1 cm are placed side by side. The difference in their lengths is

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Raj Pandey

Contributor-Level 9

(b) L₁₌ Length of rod =20.123cm

L₂₌ Length of rod = 18.1cm

Difference in length

= 20.123-18.1

=2.0 cm

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6 months ago

Overview Physics

The kinetic energy of a body varies directly as square of time elapsed. The force acting varies directly as

0 Follower 6 Views

Vishal Baghel

Contributor-Level 10

$K . E . \propto t^{2}$

$\begin{array}{l} \Rightarrow \frac{1}{2} m v^{2} \propto t^{2} \\ \Rightarrow v \propto t \\ \Rightarrow \frac{d v}{d t} = c o n s t a n t \\ \Rightarrow F = m a = \frac{m d v}{d t} = c o n s t a n t \end{array}$

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