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Chitra Varma was a great painter. On seeing his great painting, one day Srikrishna Devaraya appointed him the chief advisor. Chitra Varma accepted it unwillingly. Tenali Ramakrishna also opposed it. Thus the announcement was made in the kingdom. As Chitra Varma had no experience in handling the court matters, the people of the kingdom became impatient and they met and requested Tenali Ramakrishna for solving the problem. 

According to his plan, Ramakrishna invited the emperor to bless his son on his naming ceremony. He made all the arrangements. The emperor attended the ceremony and blessed the baby boy. Later, lunch was arranged and dishes were served. The dishes were horrible. They were not good to eat. They were not cooked properly. Then Ramakrishna brought the cook before the emperor and said that he appointed the best carpenter in the kingdom to cook dishes.

The emperor asked Ramakrishna how an excellent carpenter could become a cook. Ramakrishna replied that if a painter could be the emperor’s chief advisor, why a carpenter could not be a cook. The emperor thought and realized his mistake in appointing Chitra Varma his chief advisor. He sent Chitra Varma back to his previous job and thanked Ramakrishna. All praised Tenali Ramakrishna’s presence of mind.

Jim and Della are referred to as the Magi because first each willingly sacrificed his/her prized possession for the sake of the loved one. Then they wisely realized that unselfish love is the greatest of all gifts.

1. He was not feeling well. He also said that he was not able to walk all the way to the court.

2. Because Nasruddin did not like to appear before the judge in his old robes.

3. The neighbour

4. The neighbour

5. The neighbour’s

When he lightened their spirits he lightened his own task.

When I saw my friend with a black eye, I knew that he had been in a fight with someone.

1. redress – to set right to remedy. 

2. henpecked – being controlled by and frightened of one’s wife. 

3. black eye – an area of skin around the eye that has gone dark because it has been hit.

1. Words with prefixes : discourtesy, uncivil. 

2. Words with suffixes : instruction, reasonable.

1. words with suffixes: vanity, really. 

2. Compound words: breakfast, housemaid.

Siddharth was unable to ask his father for a cricket bat.

This is an experience I had when I was new to Mumbai. I got into a bus and asked the conductor for a ticket to Dadar. The conductor shook his head and told me that I had got into the bus going in the wrong direction. He patiently explained that I would have to get off at the next stop, cross the road, and catch a bus having the same number but going in the opposite direction. He even pointed out the bus stop to me. Though I felt a bit embarrassed, I thanked him for his kindness.

I went to the garden on my bicycle.

I studied in Aryan Primary School.

Sergeant-Major knew that the paw would cause disaster and that it had already caused enough troubles.

• During peptization a precipitate is converted into colloidal sol by shaking with dispersion medium in the presence of a small amount of an electrolyte. The electrolyte used is known as peptizing agent.
• During the process, the precipitate adsorbs one of the ions of the electrolyte on its surface and as a result, positive or negative charge is developed on the precipitate which finally breaks up into small particles of colloidal size.

[Note: This method is generally applied to convert a freshly prepared precipitate into a colloidal sol.]

i. This reaction is an example of heterogeneous catalysis. 

ii. Fe is used as a catalyst while K₂O and Al₂O₃ are promoters of the Fe catalyst. Al₂O₃ is used to prevent the fusion of Fe particles while K₂O causes chemisorption of nitrogen atoms.

Propane is an odourless and highly.

The time period of pendulum clock is given by

T = 2π\(\sqrt{\frac{l}{g}}\) , or T ∝ \(\sqrt l\) 

During summer, length of the pendulum increase due to which the time period T increase and pendulum takes more time to complete one oscillation. Thus it slows in summer.

During winter length decrease, decreasing the time period. Consequently pendulum gains time and clock becomes fast.

Area, A = 0.3 cm² = 0.3 x 10^-4 m²

Temperature, T = (2727 + 273) K = 3000 K

σ = 5.67 x 10^-5 erg cm^-2 s^-1 K^-4

= 5.67 x 10^-8 Jm^-2 s^-1 K^-4.

Power consume = Area = σT⁴

= 0.3 x 10^-4 x 0.35 x 5.67 x 10^-8 x (3000)⁴ Js^-1 or W.

= 48.22 W.

1 Molar mass M of Al has N_A = 6.023 × 10²³ atoms.  

m = mass of Al paisa coin has N=N_A m/M atoms 

Now, Z_Al = 13, M_Al = 26.9815g 

Hence N = 6.02 × 10²³ atoms/mol × 0.75 /26.9815g/mol 

= 1.6733 × 10²² atoms  

q = +ve charge in paisa = N Ze 

= (1.67 × 10²²)(13) (1.60 × 10^–19C) 

= 3.48 × 10⁴ C. 

q = 34.8 kC of ±ve charge. 

This is an enormous amount of charge. Thus we see that ordinary neutral matter contains enormous amount of ± charges.

(c) directly proportional to the density of the gas

 In the usual notation, the mean free path of a gas molecule,

\(\lambda\) = \(\frac{1}{\sqrt2\pi d^2(N/V)}\)

Density, ρ, of the gas = \(\frac{mass}{volume}\)

= \(\frac{mN}{V}\), where m is the molecular mass.

∴ \(\frac{N}{V}\) = \(\frac{ρ}{m}\)

∴ λ = \(\frac{m}{\sqrt2\pi d^2ρ}\)

The frequency of timing fork D is 340 Hz. Let n be the frequency of tuning fork C. Since tuning forks C and D produce 8 beats per second when sounded together,

n – 340 = 8 or 340 – n = 8 

∴ n = 348 Hz or 332 Hz

When the prongs of a tuning fork are filed a little, its frequency increases. Let n’ be its frequency after filing : n’ > n.

It is given that the beat frequency is reduced from 8 Hz to 4 Hz.

If n was 348 Hz, n’ will be more than 348 Hz. Hence, the beat frequency should increase. Hence, n ≠ 348 Hz.

∴ n = 332 Hz

(B) 340.8 Hz

Correct option is (A) increase

Correct option is (B) 145 Hz

(C) half the beat frequency

(A) an amplitude greater than either of the component waves

Initial speed μ = 180 kmh^-1

μ = 50 ms^-1

t = 25 s

v = u – at

v = 0

∴ 0 = 50 – a × 25

⇒ a = 2 ms^-2

v² = u² – 2as

⇒ s = μ²/2 a

(50²/(2 x 2))

∴ Stopping distance = 625 m 

Average Velocity = \(\frac{s_1+s_2}{t_1+t_2}=\frac{2\times v_1v_2}{v_1+v_2}\)

40 km/hr = \(\frac{2\times30\frac{km}{hr}\times v_2}{30\frac{km}{hr}+v_2}\)

2 = \(\frac{3v_2}{30\frac{km}{hr}+v_2}\)

60 km/hr +2v₂ = 3v₂  

v₂ = 60 km/hr 

The correct option is A) hc/ λ.

The energy of a photon of wavelength λ, is hc/ λ.

Velocity = Slope of x-t graph

V_A = tan 30° = 1√3

V_B = tan 60° = Clearly, V_B > V_A.

V_A/V_B = 1√3/√3 = 1/3 

(i) It represents that the body has uniform velocity throughout the motion.

(ii) It represents that the body has some initial velocity equal to OA and is moving with uniform acceleration.

(iii) It represents that the body has an initial velocity equal to OA and is moving with uniform retardation (negative acceleration). Example, a body thrown vertically upwards.

(iv) It represents that the body an initial velocity to OA and is moving with uniform retardation along AB. The portion BC of the graph shows that body is uniformly accelerated.

(v) AB Portion of the graph shows that the body has some initial velocity equal to OA and is moving with uniform acceleration, BC portion of the graph shows the uniform retardation of the body. CD portion shows that the body is moving with uniform velocity, DEF portion of the graph show that the velocity of the body is variable, i.e., it is accelerated and retarded.

(a) The particle was released from rest at t = 0.

(c) At C, the velocity and the acceleration vanish.

(e) The speed at D exceeds that at E.

Sage Markandeya told Yudhishtira that in every division of people, there are enlightened souls who can guide even scholars and masters of vedic teachings.

(i) We know that the magnitude of the slope of (v - t) graph in a small interval gives the magnitude of the average acceleration of the particle. It is clear from the gives figure that the magnitude of the slope is greatest in (2) and the least in (3).

Hence, the magnitude of average acceleration is the greatest in the interval (2) and least in the interval (3).

(ii) It is clear from the figure that average speed is greatest in the interval (3) and the least in the interval (1).

(iii) The speed v is positive in all the three intervals. Further, in interval (1), the slope of (v - t) graph is parallel to the time- axis, hence, a is zero in this interval.

(iv) At points A, B, C and D (v - t) graph is parallel to time-axis Therefore, 'a' is zero at all the four points.

B, represent greater velocity because it has greater slope.

Average acceleration is greatest in interval 2 Average speed is greatest in interval 3 v is positive in intervals 1, 2, and 3 a is positive in intervals 1 and 3 and negative in interval 2 a = 0 at A, B, C, D
Acceleration is given by the slope of the speed-time graph. In the given case, it is given by the slope of the speed-time graph within the given interval of time. Since the slope of the given speed-time graph is maximum in interval 2, average acceleration will be the greatest in this interval. Height of the curve from the time-axis gives the average speed of the particle. It is clear that the height is the greatest in interval 3. Hence, average speed of the particle is the greatest in interval 3.

In interval 1:
The slope of the speed-time graph is positive. Hence, acceleration is positive. Similarly, the speed of the particle is positive in this interval.
In interval 2:
The slope of the speed-time graph is negative. Hence, acceleration is negative in this interval. However, speed is positive because it is a scalar quantity. 

In interval 3:
The slope of the speed-time graph is zero. Hence, acceleration is zero in this interval. However, here the particle acquires some uniform speed. It is positive in this interval.
Points A, B, C, and D are all parallel to the time-axis. Hence, the slope is zero at these points. Therefore, at points A, B, C, and D, acceleration of the particle is zero.

If a ball is thrown up, the direction of motion of the body is the same as the direction of its velocity whereas the acceleration due to gravity acts on it in the downward direction.

Thus, the direction in which an object moves is given by the direction of velocity and not by the direction of acceleration.

Interval 3 (Greatest), Interval 2 (Least)
Positive (Intervals 1 & 2), Negative (Interval 3)
The average speed of a particle shown in the x-t graph is obtained from the slope of the graph in a particular interval of time.
It is clear from the graph that the slope is maximum and minimum restively in intervals 3 and 2 respectively. Therefore, the average speed of the particle is the greatest in interval 3 and is the least in interval 2. The sign of average velocity is positive in both intervals 1 and 2 as the slope is positive in these intervals. However, it is negative in interval 3 because the slope is negative in this interval.

Its acceleration is Zero.

This occurs when the body returns of the starting (initial) point.

Negative, Negative, Positive (at t = 0.3 s)
Positive, Positive, Negative (at t = 1.2 s)
Negative, Positive, Positive (at t = –1.2 s)

For simple harmonic motion (SHM) of a particle, acceleration (a) is given by the relation: a = – ω²x ω → angular frequency …………..… (i)
t = 0.3 s
In this time interval, x is negative. Thus, the slope of the x-t plot will also be negative. Therefore, both position and velocity are negative.
However, using equation (i), acceleration of the particle will be positive. t = 1.2 s
In this time interval, x is positive. Thus, the slope of the x-t plot will also be positive. Therefore, both position and velocity are positive. However, using equation (i), acceleration of the particle comes to be negative. t = – 1.2 s
In this time interval, x is negative. Thus, the slope of the x-t plot will also be negative. Since both x and t are negative, the velocity comes to be positive. From equation (i), it can be inferred that the acceleration of the particle will be positive.

At the triple point of water,

T = 273.16 K and R = 101.6 Ω (given)

∴ R = R₀[1 + 5 x 10^-3(T - T₀)]

⇒ 101.6 = R₀[1 + 5 x 10^-3 (273.16 - T₀)]   ...(i)

At temperature, T = 600.5 K, R = 165.5 Ω

∴ 165.5 = R₀[1 + 5 x 10^-3(600.5 - T₀)]  ...(ii)

Dividing (ii) by (i), we have,

165.5/101.6 = {R₀[1 + 5 x 10^-3(600.5 - T₀)]}/{[R₀[1 + 5 x 10^-3(273.16 - T₀)]]}

⇒ 1.63[1 + 5 x 10^-3(273.16 - T₀)]

= [1 + 5 x 10^-3(600.5 - T₀)]

⇒ 1.63 x 5 x 10^-3(273.16 - T₀)] + 1.63

= 1 + 5 x 10^-3(600.5 - T₀)

⇒ 5 x 10^-3(600.5 - T₀ - 1.63 x 273.16 + 1.63 T₀) = 1.63 - 1

∴ 0.63 T₀ + 155.25 = {0.63}/{5 x 10^-3}

⇒ T₀ = -46.42 K   ....(iii)

∴ From equation (i),

101.6 = R₀[1 + 5 x 10^-3{273.16 - (-46.42)}]

⇒ R₀ = 101.6/2.6 = 39.1 Ω   ...(iv)

Now, when R = 123.4 Ω, T = ?

∴ Using, R = R₀[1 + 5 x 10^-3(T - T₀)]

Taking value of R₀ and T₀ from equations (iii) and (iv), we get

123.4 = 39.1 x [1 + 5 x 10^-3{T - (-46.42)}]

or, 1 + 5 x 10^-3(T + 46.42) = 123.4/39.1 = 3.156

or, T + 46.42 = {3.156 - 1}/{5 x 10^-3} = {2.156 x 1000}/{5} = 431.2

∴ T = 431.2 - 46.42 = 384.78 K

Area under velocity-time graph indicates the displacement.

u + v = 6 and u - v = 4 ⇒ 2u = 10

or, u = 5 ms^-1 2v = 6 - 4 = 2 ⇒ v = 1 ms^-1.

In the S.H.M., acceleration, a = -ω²x, where ω (i,e., angular frequency) is constant.

(i) At time t = 0.3s, x is negative, the slope of x - t plot is negative hence position and velocity are negative. Since a = -ω²x, hence acceleration is positive.

(ii) At time t = 1.2 s, x is positive, the slope of x -1 plot is also positive, hence position and velocity are positive. Since, - ω²x, hence acceleration is negative.

(iii) At t = -1.2 s, x is negative, the slope of x - t plot is also negative. But since both x and t are negative here, hence velocity is positive. Finally acceleration 'a' is also positive.

Area of a-t graph represents the change in velocity of the body in a given time interval.

Since acceleration is the force on unit mass, the acceleration due to gravity is constant.