Class 11th

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

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Raj Pandey

Contributor-Level 9

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

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

$tan θ = 1$

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3 months ago

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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Class 11th Work, Energy and Power

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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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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Class 11th Physics Motion in Plane

Contributor-Level 10

Work done = $Δ K E$

$\Rightarrow - f = - x$

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

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3 months ago

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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Contributor-Level 10

Time (minimum) taken to cross the river

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

$100 * 10^{- 3} h$

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

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3 months ago

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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3 months ago

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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Contributor-Level 10

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

Component of weight along incline.

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

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

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3 months ago

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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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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3 months ago

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

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Class 11th Physics Oscillations

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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3 months ago

In the diagram shown below, the block has charge $Q$ and is attached at one end of the light spring. The surface is smooth and horizontal and spring is in its natural length. If the electric field $E$ is switched on, the maximum elongation in the spring will be

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Contributor-Level 10

From work energy theorem

$W_{s p} + W_{E} = ? K$

$- \frac{1}{2} k x_{0}^{2} + Q E . x_{0} = 0$

$? x_{0} = \frac{2 Q E}{k}$

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3 months ago

A liquid of coefficient of cubical expansion $γ$ is filled in a container having coefficient of linear expansion $\frac{γ}{3} .$ On heating, liquid level in the container will

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