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9 months ago

Physics physics ncert exemplar solutions class 12th chapter two

A geyser heats water flowing at a rate of 2.0kg per minute from 30°C to 70°C. If geyser operates on a gas burner, the rate of combustion of fuel will be________g min^-1.

[Heat of combustion = 8 * 10³ Jg^-1,

Specific heat of water = 4.2Jg^-1 °C^-1]

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Vishal Baghel

Contributor-Level 10

Heat given to water by geyser

$H = {\overset{o}{m}}_{ω} C Δ T$

$= 2 * 1 0^{3} \frac{g r a m}{m i n} * 4.2 J g^{- 1} c^{- 1} * (70 - 30) ° C$

$= 2 * 42 * 4 * 1 0^{3} J / m i n$ - (1)

heat provided by combustion of fuel

$H = {\overset{o}{m}}_{f} C_{f}$

${\overset{o}{m}}_{f} = \frac{2 * 42 * 4}{8} = 42$

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9 months ago

Physics physics ncert exemplar solutions class 12th chapter two

A system to 10 balls each of mass 2kg are connected via massless and unstretchable string. The system is allowed to slip over the edge of a smooth in figure. Tension on the string between the 4^th and 8^th ball is_________N when 6^th ball just leaves the table.

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Vishal Baghel

Contributor-Level 10

12 g – T = 12 a- (1)

T – T' = 2a- (2)

T' = 6a- (3)

From (1) & (2)

12g – 12a – T' = 2a- (4)

From (4) & (3)

12g – 12a – T' = 14

$12 g = \frac{10}{3} T' \Rightarrow T' = \frac{120 * 3}{10} = 36$

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9 months ago

Physics Work, Energy and Power

A fighter jet is flying horizontally at a certain with a speed of 200ms^-¹. When it passes directly overhead ad anti-aircraft gun, a bullet is fired from the gun, at an angle q with the horizontal, to hit the jet. If the bullet speed is 400m/s, the value of q will be _____________.

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Payal Gupta

Contributor-Level 10

B fighter jet

A anti-Air craft gun

Draw velocity diagram of A w.r.t. B

If A hits B

Then Relative velocity perpendicular to the line joining A to B will be zero.

That means 400 cos = 200

$\begin{array}{l} c o s θ = \frac{1}{2} \\ θ = 60 ° \end{array}$

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9 months ago

Physics physics ncert exemplar solutions class 12th chapter two

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New answer posted

9 months ago

Physics physics ncert exemplar solutions class 12th chapter two

A batsman hits back a back a ball of mass 0.4 kg straight in the direction of the bowler without changing its initial speed of 15ms^-1. The impulse imparted to the ball is__________Ns.

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Vishal Baghel

Contributor-Level 10

Impulse = change in momentum

= $Δ P = 2 m v$

= 2 * 15 * 0.4

= 12 N sec

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9 months ago

Physics physics ncert exemplar solutions class 12th chapter two

A ball is projected vertically upward with an initial velocity of 50ms^-1 at t = 0s. At t =2 2s, another ball is projected vertically upward with same velocity. At t = _______s, second ball will meet the first ball (g = 10 ms^-2).

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Vishal Baghel

Contributor-Level 10

$h = 50 t - \frac{1}{2} ? * 10 t^{2} = 80 m$

Now velocity B with respect to A

$V_{B A} = V_{B} - V_{A}$

= 50 – 30

= 20 m/sec

Now $A_{B A} = a_{B} - a_{A}$

= $10 (- \hat{j}) - (- 10 \hat{j})$

= 0

Time taken to collide.

$h = V_{B A} t$

80 = 20t

t = 4 sec

Total time = 4 + 2

= 6 sec

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9 months ago

Physics Communication Systems

Choose the correct statement for amplitude modulation

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Payal Gupta

Contributor-Level 10

In amplitude modulation, the amplitude of the career signal is varied in according with the modulating signal.

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9 months ago

Physics physics ncert exemplar solutions class 12th chapter two

A sinusoidal wave y(t) = 40 sin (10 * 10⁶pt) is amplitude modulated by another sinusoidal wave x(t) = 20sin(1000πt). The amplitude of minimum frequency component of modulated signal is:

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Vishal Baghel

Contributor-Level 10

$C_{m} (t) = A_{C} s i n ω_{C} t + \frac{μ A_{C}}{2} [c o s (ω_{C} - ω_{m}) t - c o s (ω_{C} + ω_{m}) t]$

$μ = \frac{A m}{A_{C}}$

$A_{C} = 40$

$μ = \frac{1}{2}$

So amplitude of minimum frequency = $\frac{μ A_{C}}{2}$

= 10

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9 months ago

Physics Moving Charges and Magnetism

The I-V characteristics of p-n junction diode in forward bias is shown in the figure. The ration of dynamic resistance, corresponding to forward bias voltage of 2 V and 4 V respectively, is

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Payal Gupta

Contributor-Level 10

Dynamic Resistance at 2v

= $\frac{Δ v}{Δ I} = \frac{2.1 - 2}{(10 - 5) * 1 0^{- 3}} = \frac{0.1}{5} * 1 0^{3}$

Dynamic Resistance at 4 v

= $\frac{Δ v}{Δ I} = \frac{4.2 - 4}{(250 - 200)} = \frac{0.2}{50 * 1 0^{- 3}} = \frac{0.2}{50} * 1 0^{3}$

$\frac{R_{2 v}}{R_{4 v}} = \frac{0.1 * 1 0^{3} * 50}{5 * 0.2 * 1 0^{3}} = 5 : 1$

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9 months ago

Physics physics ncert exemplar solutions class 12th chapter two

The positive feedback is required by an amplifier to act an oscillator. The feedback here means:

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Vishal Baghel

Contributor-Level 10

A portion of the output power is returned back to the input.

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