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We know that the SP of 10 bananas = ₹36

SP of 1 banana = ₹36/10 = ₹3.6

SP of 1 dozen bananas = ₹3.6 × 12 = ₹43.20

CP of 1 dozen bananas = ₹40

Since, SP is more than CP so it’s a gain.

Gain = SP – CP = 43.20 – 40 = ₹3.2

Gain % = (3.2 × 100) / 40

=8%

∴ The Gain percent is 8%

The substance synthesized in cell responsible for increasing osmotic is Glucose.

Correct option is B) decreases

Pressure is applied on ice with skating boots. As the pressure increases melting point ice decreases. So ice melts at a lower temperature than 0°C. Therefore skating is possible on ice.

Inside the matter, we observe the increase/decrease in volume during the change of state.

Correct option is A) liquid

Correct option is C) humidity

Correct option is A) A

The pressure exerted due to osmosis is osmotic pressure now termed as osmotic potential.

Water when cooled in a refrigerator, it becomes ice. (Decreasing the temperature).

The passive absorption of solutes when mediated by a carrier is called facilitated diffusion.

The molecules in the gas are far from each other i.e., the inter molecular space is very high than in solids and liquids. Hence they have less attractive power. So, gases diffuse faster than solids and liquids.

Smoke of incense stick is a gas. Gas diffuses in air only. So the smoke cannot diffuse in the absense of air.

The process of diffusion is responsible for spreading the fragrance of an incense stick in one corner of room.

(a) Diffusion is the process involved in this as the fragrance quickly spreads in the entire room. 

(b) The smell of food being cooked reaches the other room by the diffusion of gases released into the air during the cooking of food.

This illustration shows that the particles are constantly moving in all directions.

The smell from deodorant spray reaches someone on the other end faster than the smell from incense stick.

Yes, it will be seen.

दिया है, q = 1.0 μC =1.0 x 10^-6 C, 

l = 2.0 मिमी = 2 x 10^-3 मी 

विद्युत द्विध्रुव आघूर्ण, p = q . 2l = 1.0 x 10^-6 x 2 x 2 x 10^-3 

= 4 x 10^-9 कूलॉम-मी

इलेक्ट्रॉन का त्वरण अधिक होगा, क्योंकि प्रोटॉन की अपेक्षा इलेक्ट्रॉन का द्रव्यमान कम होता है।

कण पर आवेश की प्रकृति के अनुसार उस पर वैद्युत-क्षेत्र ऐसी दिशा में लगाकर, ताकि उसके कारण कण पर लगने वाला वैद्युत बल ऊर्ध्वाधरत: ऊपर की ओर अर्थात् कण के भार की विपरीत दिशा में कार्य करे तथा परिमाण में इसके बराबर हो।

आवेशित खोखले गोलाकार चालक के भीतर वैद्युत क्षेत्र की तीव्रता शून्य होती है|

Principle of multiplication of thrust by transmission of fluid pressure : The normal force exerted by a fluid on any surface in contact with it is called the thrust. Consider two hydraulically connected cylinders, one of cross section a and the other A, as in figure. If a force F_a is exerted on the smaller piston, pressure p = \(\frac{F_a}a\) s produced and transmitted undiminished throughout the liquid. Then, the thrust F_A on the larger piston is

F_A = p_A = \(\frac{A}a\)F_a

If A = na, F_A = nF_a,i.e., the thrust on the larger piston is multiplied n times. This is known as the principle of multiplication of thrust by transmission of fluid pressure.

Carburetor is that chamber of an internal combustion engines in which air is mixed with petrol. Air is allowed to enter through a nozzle with high velocity. The pressure is lowered at that pint and petrol is sucked up into the chamber the petrol vaporizes quickly. The mixed vapours of petrol and air are then fed into the cylinder of internal combustion engine.

h ∝ 1/r, Capillary rise will be halved.

The height upto which a liquid rises in capillary tube, is given by:-

h = 2scosθ/rγg

Now, as the surface tension(s) of hot water is less than that of cold water, when a capillary tube is immersed in hot water, it expands. Therefore, its radius increases. As 'δ' decreases and 'r' increases. therefore the height of capillary size(h) will be smaller in hot water than in cold water.

Correct option is  B) ∠i = ∠c

Mass of the block, m₁ = 0.20 kg

= 0.02 x 1000 = 200 gm

Temperature of the block, θ₁ = 150°C

Mass of water in the calorimeter,

m₂ = Volume x density = 150 x 1g = 150g

Water equivalent,

W = 0.025 Kg = 0.025 x 1000 = 25 kg

Temperature of calorimeter, θ₂ = 27°C

Specific heat capacity of water,

C₂ = 1 cal g^-1 = 4.184 Jg^-1

Let C₁ be the specific heat capacity of the block. Let θ be the final temperature. Then θ = 40°C. Using the principle of calorimetry, we get

m₁C₁ (θ₁ - θ) = (m₁ + ω) C₂(θ - θ₂)

⇒ C₁ = {(m₂ + ω) C₂(θ - θ₂)}/{m₁(θ₁ - θ)}

= {(150 + 25) x 4.184 x (40 - 27)}/{200 (150 - 40)}

= 0.43 Jg^-1K^-1

If the heat losses to the surrounding are not negligible, the answer will be smaller compared to 0.43 Jg^-1K^-1.

1. In method of mixtures, solid sample gives away heat to liquid, and heat exchange between, solid, liquid and calorimeter is considered as isolated.

2. If density of solid sample is lesser than liquid in calorimeter, sample will float on liquid.

3. This will cause partial heat loss to air inside the calorimeter and standard heat equations of calorimeter will no longer be applicable.

Hence, in method of mixtures it is essential that density of solid sample be greater than liquid in calorimeter.

एक से 100 तक की गिनती में 8 बीस बार आएगा।

Rising of coin when water is poured in a cylindrical transparent vessel.

D) a → s,b → r,c → q,d → p

It is the ratio of speed of light in vacuum to speed of light in medium.

n = c/v

C) c = sin^-1(3/4) 

1. When does a mirage form? 

2. How does a mirage form?

Correct option is  A) c/v

(a) False: It is given that the motion of the particle is arbitrary. Therefore, the average velocity of the particle cannot be given by this equation.

Absolute refractive index of water = 4/3

We know that sin C = 1/Refractive index 

Sin C = 1/4/3 = sin C = 3/4 = 0.75

Critical angle of water = C = 48°5′ = 48.5°.

(b) True: The arbitrary motion of the particle can be represented by this equation.

The angle of incidence at which the light ray propagates from denser to rarer graze along interface is called critical angle of denser medium.

Since the motion is arbitrary, the acceleration may not be uniform. Therefore, relations (c) and (d) cannot be correct, for the acceleration in these equations cannot be constant.

Again for an arbitrary motion, the average velocity cannot be defined as in equation (a).Therefor, the relation (a) is also not correct.

Hence, only relations (b) and (e) are correct relation.

Stress = K × strain

= K(ΔV/V)

=K(3α)Δt

= 140 x 10⁹ x 3 x 1.7 x 10^-5 x 20

= 1.428 x 10⁸ N/m² 

This is about 10³ times atmospheric pressure

Conduction rate (P_cond) is the amount of energy transferred per unit time through a slab of area A and thickness x, the two sides of the slab being at temperatures T₁, and T₂ (T₁ > T₂), and is given P_cond = \(\frac{Q}{t}\) = kA \(\frac{(T_1-T_2)}{\text{x}}\)

1) The ball shines. 

2) The ball appears to raise up in water.

(a) matches with (iv)

(b) matches with (iii)

(c) matches with (i)

(d) matches with (ii)

(a) matches with (iv)

(b) matches with (iii)

(c) matches with (i)

(d) matches with (ii)

(a) matches with (ii)

(b) matches with (i)

(c) matches with (iv)

(d) matches with (iii)

1. Consider a metallic rod of length l₀ fixed between two rigid supports at T°C. 

If the temperature of rod is increased by ∆T, length of rod would become, l = l₀(1 + α∆T)

Where, α is the coefficient of linear expansion of material of the rod.

But the supports prevent expansion of rod. As a result, rod exerts stress on the supports. Such stress is termed as thermal stress.

2. Disadvantage: Thermal stress can lead to fracture or deformation in substance under certain conditions. 

3. Railway tracks are made up of metals which expand upon heating. If no gap is kept between tracks, in hot weather, expansion of metal tracks may exert thermal stress on track. This may lead to bending of tracks which would be dangerous. Hence, railway track is not a continuous piece but is made up of segments separated by gaps.

(i) When the balls hits the ground; their vertical velocities are equal. However the horizontal velocities will be different. Hence the resultant velocities of the two balls are different, so the balls would hit the ground with different velocities.

(ii) Both the balls would reach the ground at the same time because their initial vertical velocities, acceleration and distance covered are all equal.

(a) -- (iv)

(b) -- (iii)

(c) -- (i)

(d) -- (ii)