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Physics Laws of Motion

A ball of mass 0.5 kg is dropped from the height of 10m. The height, at which the magnitude of velocity becomes equal to the magnitude of acceleration due to gravity, is __________m.

[Use g = 10m/s²]

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Payal Gupta

Contributor-Level 10

Let 'h' be the height at which velocity becomes equal to magnitude of Acceleration

v = g = 10

v = u + at

10 = 0 + 10t

t = 1 sec

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

$= 0 * 1 + \frac{1}{2} * 10 * 1 * 1$

= 5m

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a year 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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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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a year 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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a year ago

Physics physics ncert exemplar solutions class 12th chapter two

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New answer posted

a year 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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a year 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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a year 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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a year 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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New answer posted

a year 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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