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Physics NCERT Exemplar Solutions Class 12th Chapter Three

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Physics NCERT Exemplar Solutions Class 12th Chapter Three Class 12th

The variation of applied potential and flowing through a given wire is shown in figure. The length of wire is 31.4cm. The diameter of wire is measured as 2.4cm. The resistivity of the given wire is measured as x * 10^-3 $Ω c m .$ The value of x is_________.

[Take p = 3.14]

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alok kumar singh

Contributor-Level 10

R = tanq = tan 45° = $\frac{ρ L}{A}$

$? 1 = ρ * \frac{31.4}{π {(1.2)}^{2}}$

$ρ = \frac{3.14 * 1.2 * 1.2}{31.4}$

$ρ = 12 * 12 * 1 0^{- 3}$

$ρ = 144 * 1 0^{- 3} = x * 1 0^{- 3} Ω - c m$

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Physics NCERT Exemplar Solutions Class 12th Chapter Three Class 12th

Two coil require 20 minutes and 60 minutes respectively to produce same amount of heat energy when connected separately to the same source. If they are connected in parallel arrangement to the same source; the time required to produce same amount of heat by the combination of coils, will be_________ min.

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alok kumar singh

Contributor-Level 10

$H_{1} = \frac{v^{2}}{R_{1}} t_{1} = \frac{v^{2}}{R_{1}} * 20$

H2 $= \frac{v^{2}}{R_{2}} * t_{2} = \frac{v^{2}}{R_{2}} * 60$

from question,

H1 = H2

$\frac{v^{2}}{R_{1}} * 20 = \frac{v^{2}}{R_{2}} * 60$

$\frac{R_{2}}{R_{1}} = 3$

Now connected is parallel

Req = $\frac{R_{1} R_{2}}{R_{1} + R_{2}}$

H = $(\frac{v^{2}}{R_{1}} + \frac{v^{2}}{R_{2}}) t'$

H = $v^{2} [\frac{R_{1} + R_{2}}{R_{1} R_{2}}] t'$

Now, According to question

H = H1 + H2

$v^{2} (\frac{R_{1} + R_{2}}{R_{1} R_{2}}) t' = \frac{v^{2}}{R_{1}} * 20$

$t' [1 + \frac{\frac{R_{2}}{R_{1}}}{R_{2}}] = \frac{20}{R_{1}}$

$t' [1 + 3] = 20 * 3$

$t' = \frac{60}{4} = 15 m i n$

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Physics NCERT Exemplar Solutions Class 12th Chapter Three Motion in a Straight Line

Two balls A and B placed at the top of 180m tall tower. Ball A is from the top at t = 0s. Ball B is thrown vertically down with an initial velocity 'u' at t = 2s. After a certain time, both balls meet 100m above the ground. Find the value of 'u' in ms^-¹.

[use g = 10 ms^-²] :

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alok kumar singh

Contributor-Level 10

Let 't' be the time taken by B to meat A.

for A, u = 0

t = [2 + t]

h = 80m

a = 10 m/sec²

h = ut + $\frac{1}{2} a t^{2}$

80 = 0 * t + $\frac{1}{2} (10) {(2 + t)}^{2}$

16 = (2 + t)²

t + 2 = 4

t = 2sec

Now for B

h = ut + $\frac{1}{2} a t^{2}$

...more

Let 't' be the time taken by B to meat A.

for A, u = 0

t = [2 + t]

h = 80m

a = 10 m/sec²

h = ut + $\frac{1}{2} a t^{2}$

80 = 0 * t + $\frac{1}{2} (10) {(2 + t)}^{2}$

16 = (2 + t)²

t + 2 = 4

t = 2sec

Now for B

h = ut + $\frac{1}{2} a t^{2}$

$? 80 = u * 2 + \frac{1}{2} * 10 t^{2}$

80 = 2u + 5 * 2 * 2

80 = 2u + 20

2u = 60

u = 30m/sec

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Physics NCERT Exemplar Solutions Class 12th Chapter Three Motion in a Straight Line

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Payal Gupta

Contributor-Level 10

Let 't' be the time taken by B to meat

for A, u = 0

t = [2 + t]

h = 80m

a = 10 m/sec²

h = ut + $\frac{1}{2} a t^{2}$

80 = 0 * t + $\frac{1}{2} (10) {(2 + t)}^{2}$

16 = (2 + t)²

t + 2 = 4

t = 2sec

Now for B

h = ut + $\frac{1}{2} a t^{2}$

$\Rightarrow 80 = u * 2 + \frac{1}{2} * 10 t^{2}$

80 = 2u + 5 * 2 * 2

80 = 2u + 20

2u = 60

u = 30m/sec

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Physics NCERT Exemplar Solutions Class 12th Chapter Three Class 12th

A potentiometer wire of length 300 cm is connected in series with a resistance 780 $Ω$ and a standard cell of emf 4V. A constant current flows through potentiometer wire. The length of the null point for cell of emf 20 mV is found to be 60 cm. The resistance of the potentiometer wire is ________ $Ω$ .

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alok kumar singh

Contributor-Level 10

$I = \frac{4}{R + 780}$

Potential Difference across AB = l * R = $\frac{4 R}{R + 780}$

Potential Difference across AC = $(\frac{4 R}{R + 780}) * \frac{60}{300}$

$\Rightarrow \frac{4 R}{5 (R + 780)} = 2 * 1 0^{- 2} \Rightarrow R = 20 Ω$

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Physics NCERT Exemplar Solutions Class 12th Chapter Three Class 12th

Resistances are connected in a meter bridge circuit as shown in the figure. The balancing length $l_{2}$ is 40cm. Now as unknown resistance x is connected in series with P and new balancing length is found to be 80cm measured from the same end. Then the value of x will be _________ $Ω.$

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Vishal Baghel

Contributor-Level 10

Initially, $\frac{P}{Q} = \frac{40}{60} = \frac{2}{3}$ - (1)

Finally, $\frac{P + x}{Q} = \frac{80}{20} = \frac{4}{1}$

(2) (1)

$\frac{P + x}{P} = \frac{4 * 3}{2} = 6$

$\Rightarrow 1 + \frac{x}{P} = 6$

$\frac{x}{P} = 5$

x = 5 P = 5 x 4 = 20 $Ω$

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Physics NCERT Exemplar Solutions Class 12th Chapter Three Class 12th

A battery of 6V is connected to the circuit as shown below. The current I drawn from the battery is:

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alok kumar singh

Contributor-Level 10

The circuit is balanced whetstone bridge.

$\Rightarrow R_{e q} = \frac{6 * 12}{6 + 12} + 2 = 6 Ω$

$\Rightarrow I = \frac{V}{R_{e q}} = \frac{6}{6} = 1 A$

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Physics NCERT Exemplar Solutions Class 12th Chapter Three Class 12th

The current I in the above given circuit will be:

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Vishal Baghel

Contributor-Level 10

$R = \frac{8}{2} Ω = 4 Ω$ (wheat stone bridge)

$I = \frac{V}{R} = \frac{40}{4} = 10 A$

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Physics NCERT Exemplar Solutions Class 12th Chapter Three Motion in a Straight Line

A scooter accelerates from rest for time t₁ at constant rate a₁ and then retards at constant rate a₂ for time t₂ and comes to rest. The correct value of $\frac{t_{1}}{t_{2}}$ will be

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Payal Gupta

Contributor-Level 10

$t a n α = \frac{v_{m a x}}{t_{1}} = a_{1}$

$t a n β = \frac{v_{m a x}}{t_{2}} = a_{2}$

$\Rightarrow \frac{a_{1}}{a_{2}} = \frac{t_{2}}{t_{1}} \Rightarrow \frac{t_{1}}{t_{2}} = \frac{a_{2}}{a_{1}}$

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Physics NCERT Exemplar Solutions Class 12th Chapter Three Motion in a Straight Line

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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