Tags physics ncert exemplar solutions class 12th chapter two

physics ncert exemplar solutions class 12th chapter two

Get insights from 123 questions on physics ncert exemplar solutions class 12th chapter two, answered by students, alumni, and experts. You may also ask and answer any question you like about physics ncert exemplar solutions class 12th chapter two

Follow Ask Question

123

Questions

Discussions

Active Users

Followers

New question posted

2 months ago

physics ncert exemplar solutions class 12th chapter two Dual Nature of Radiation and Matter

A heat engine operates with the could reservoir at temperature 324K. The minimum temperature of the hot reservoir, if the heat engine takes 300J heat from the hot reservoir and delivers 180J heat to the cold reservoir per cycle, is__________K.

0 Follower 2 Views

New answer posted

2 months ago

physics ncert exemplar solutions class 12th chapter two Class 12th

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]

0 Follower 3 Views

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

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

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

0 0

New answer posted

2 months ago

physics ncert exemplar solutions class 12th chapter two Work, Energy and Power

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.

0 Follower 4 Views

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

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

0 0

New question posted

2 months ago

physics ncert exemplar solutions class 12th chapter two Class 12th

0 Follower 3 Views

New answer posted

2 months ago

physics ncert exemplar solutions class 12th chapter two Electrostatic Potential and Capacitance

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.

0 Follower 2 Views

Vishal Baghel

Contributor-Level 10

Impulse = change in momentum

= $Δ P = 2 m v$

= 2 * 15 * 0.4

= 12 N sec

0 0

New answer posted

2 months ago

physics ncert exemplar solutions class 12th chapter two Units and Measurement

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).

0 Follower 2 Views

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

0 0

New answer posted

2 months ago

physics ncert exemplar solutions class 12th chapter two Communication Systems

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:

0 Follower 2 Views

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

0 0

New answer posted

2 months ago

physics ncert exemplar solutions class 12th chapter two Semiconductor Electronics

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

0 Follower 1 View

Vishal Baghel

Contributor-Level 10

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

0 0

New answer posted

2 months ago

physics ncert exemplar solutions class 12th chapter two Nuclei

A radioactive nucleus can decay by two different processes. Half-life for the first process is 3.0 hours while it is 4.5 hours for the second process. The effective half-life of the nucleus will be:

0 Follower 1 View

Vishal Baghel

Contributor-Level 10

$\frac{1}{t} = \frac{1}{t_{1}} + \frac{1}{t_{2}}$

$\frac{1}{t} = \frac{1}{3} = \frac{1}{4.5}$

$= \frac{4.5 + 3}{3 * 4.5}$

t = $\frac{3 * 4.5}{7.5} = \frac{9}{5}$

t = 1.8 hrs

0 0

New answer posted

2 months ago

physics ncert exemplar solutions class 12th chapter two Dual Nature of Radiation and Matter

A metal surface is illuminated by a radiation of wavelength 4500 . The ejected photo-electron enters a constant magnetic field of 2 mT making an angle of 90° with the magnetic field. If it starts revolving in a circular path of radius 2nm, the work function of the metal is approximately:

0 Follower 2 Views

Vishal Baghel

Contributor-Level 10

$k = \frac{h c}{λ} - ?$

As radius, r = $\frac{m v}{e B} = \frac{\sqrt[]{2 k m}}{e B}$

$\begin{array}{l} \Rightarrow {(r e B)}^{2} = 2 k m \\ \Rightarrow ? = \frac{h c}{λ} - \frac{{(r e B)}^{2}}{2 m} \end{array}$

0 0

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

65k Colleges
1.2k Exams
688k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Need guidance on career and education? Ask our experts

Your Question

physics ncert exemplar solutions class 12th chapter two

Questions

Discussions

Active Users

Followers

All

Discussions

Unanswered Questions

Share Your College Life Experience