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6 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Three

A Tennis ball is released from a height $h$ and after freely falling on a wooden floor it rebounds and reaches height $\frac{h}{2}$ . The velocity versus height of the ball during its motion may be represented graphically by:

(graphs are drawn schematically and on not to scale)

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alok kumar singh

Contributor-Level 10

$\begin{array}{r} \Rightarrow V^{2} = U^{2} + 2 g S \\ \Rightarrow V^{2} = 0 + 2 g (h - y) \\ \Rightarrow V^{2} = 2 g h - 2 g y \\ \Rightarrow V = \sqrt[]{2 g h - 2 g y} \end{array}$

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6 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Nine

In experiment of find acceleration due to gravity (g) using simple pendulum, time period of 0.5s is measured from time of 100 oscillation with a watch of 1 s resolution. If measured value of length is 10 cm known to 1 mm accuracy, The accuracy in the determination of g is found to be x%. The value of x is_______.

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Vishal Baghel

Contributor-Level 10

T = 0.5 sec

No. of oscillation = 100

Resolution = 1 sec

$l = 10 c m \pm . 1 c m$

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6 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Nine

In an experiment with a convex lens, The plot of the image distance (v') against the object distance $(μ')$ measured from the focus gives a curve v $' μ'$ = 225. If all the distances are measured in cm. The magnitude of the focal length of the lens is__________cm.

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Vishal Baghel

Contributor-Level 10

As, $\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{v' + f} - \frac{1}{μ' - f} = \frac{1}{f}$

Also, $μ' v' = 225$

Using Newton's formula - $μ' v' = f^{2}$

So, f² = 225

f = 15 cm

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6 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Nine

An object 'O' is placed at a distance of 100 cm in front of a concave mirror of radius of curvature 200 cm as shown in the figure. The object starts moving towards the mirror at a speed 2 cm/s. The position of the image from the mirror after 10 s will be at________cm.

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Vishal Baghel

Contributor-Level 10

u = 100 cm

$f = \frac{R}{2} = - 100 c m$

after 10 sec $i = \frac{6}{6} = 1 A$

u = 80 m

$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{- 100} - \frac{1}{- 80} = \frac{1}{80} - \frac{1}{100} = \frac{5 - 4}{400} = \frac{1}{400}$

$∴ v = 400 c m$

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6 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Three

For the given circuit the current through battery of 6 V just after closing the switch 'S' will be_______A.

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Vishal Baghel

Contributor-Level 10

Kindly consider the following answer

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6 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Three

For the given circuit the current through battery of 6 V just after closing the switch ‘S’ will be_______A.

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6 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Three

An electrical bulb rated 220 V, 100 W, is connected in series with another bulb rated 220 V, 60 W. If the voltage across combination is 220 V, the power consumed by the 100 W bulb will be about ______W.

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Vishal Baghel

Contributor-Level 10

$\frac{v_{1}}{v_{2}} = \frac{R_{1}}{R_{2}} = \frac{P_{2}}{P_{1}} = \frac{60}{100} = \frac{3}{5}$

$v_{1} = \frac{3}{8} * 220 = 3 * 27.5 = 82.5 v$

$∴ P_{1} = {(\frac{v_{1}}{v})}^{2} P$

= 14 w

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6 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Fourteen

The potential energy of a particle of mass 4 kg in motion along the x-axis is given by $U = 4 (1 - c o s 4 x) J .$ The time period of the particle for small oscillation (sin = ) is $(\frac{π}{K}) s .$ The value of K is_______.

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Vishal Baghel

Contributor-Level 10

m = 4kg

U = 4 (1 – cos 4x) J

$F = - \frac{\partial υ}{\partial x} = - 4 * 4 s i n 4 x$

$∴ a = - \frac{16}{4} s i n 4 x = - 4 s i n 4 x = - 16 x$

$\Rightarrow a = - 16 x$

$ω^{2} = 16 \Rightarrow ω = 4 = \frac{2 π}{T} \Rightarrow T = \frac{2 π}{4} = \frac{π}{2} s e c$

k = 2

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6 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Thirteen

At a certain temperature, the degrees of freedom per molecule for gas is 8. The gas performs 150 J at work when it expands under constant pressure. The amount of heat absorbed by the gas will be________J.

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Vishal Baghel

Contributor-Level 10

Degree of freedom, f = 8

$∴ C_{V} = \frac{8}{2} R = 4 R a n d C_{P} = 4 R + R = 5 R$

So, H = $Δ U + W$

at constant pressure, W = $P Δ v$

150 = $n R Δ T$

So, $H = n C_{V} Δ T = n 5 R Δ T = 5 (n R Δ T)$

= 5 * 150

= 750 J

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6 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Nine

A string of area of cross – section 4 mm² and length 0.5 m is connected with a rigid body of mass 2 kg. The body is rotated in a vertical circular path of radius 0.5m. The body acquires a speed of 5 m/s at the bottom of the circular path. Strain produce in the string when the body is at bottom of the circle is________* 10^-5.

(use young's modulus 10¹¹ N/m² and g = 10 m/s²)

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Vishal Baghel

Contributor-Level 10

For circular motion:-

$T - m g = \frac{m v^{2}}{r}$

$\Rightarrow T = m (g + \frac{v^{2}}{r})$

$= 2 (10 + \frac{25}{1 / 2}) = 120 N$

$\Rightarrow S t r a i n = \frac{s t r e s}{Y}$

= 30 * 10^-5

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