# Physics GS TIFR- Tata Institute of Fundamental Research Questions 6-10 _ Solved

6. Evidence for the expansion of the universe comes from the observation of light from distant galaxies. What is this observation?

**(i) Light from distant objects is red shifted**

(ii) More distant pulsars have a lower orbital frequency

(iii) Light from other galaxies is focused by gravitational lensing

(iv) The number of stars per volume drops at greater distances

Answer : The observation that provides evidence for the expansion of the universe is known as the redshift of light from distant galaxies. When light is emitted from an object, it has a certain wavelength and frequency. As the light travels through space, its wavelength can be stretched or compressed, depending on the relative motion of the object emitting the light and the observer receiving it. This effect is known as the Doppler shift.

In the case of light from distant galaxies, the expansion of the universe causes the galaxies to move away from us, and as a result, the light they emit is shifted towards longer wavelengths. This shift is called redshift because it shifts the light towards the red end of the spectrum. The amount of redshift is proportional to the distance between us and the galaxy, as well as the expansion rate of the universe.

By measuring the redshift of light from distant galaxies, astronomers can determine how fast they are moving away from us, and therefore, infer the rate of expansion of the universe. The observation of redshifted light from distant galaxies provides strong evidence that the universe is expanding.

7. The wheels on a drag racing car have a higher diameter at the end of a race than at the beginning. Which of the following is the least important factor contributing to this?

(i) The weight of the car is reduced due to fuel consumption.

(ii) The tires heat up so the air inside expands.

(iii) The tires rotate rapidly so the centrifugal force stretches them.

**(iv) Due to rubber fatigue the stiffness of the rubber decreases.**

Answer: Assuming the wheels on the drag racing car do indeed have a higher diameter at the end of the race than, at the beginning, the least important factor contributing to this change in diameter would likely be the color of the car's paint. The change in diameter could be due to a number of factors, such as:

The heat generated during the race causes the tires to expand

The wear and tear of the tires during the race leading to a reduction in tire pressure and therefore an increase in diameter

The effects of the track surface, such as debris or rubber buildup, on the tires.

While the color of the car's paint may have some effect on the temperature of the tires due to its ability to absorb or reflect sunlight, it would likely be a very small factor compared to the other potential causes listed above.

8. 11 balls have radii X cm, X+1 cm, Ã‰, X+10 cm. Which of the following is true about the volumes of these balls?

The mean is equal to the median.

The mean is greater than the median.

The mean is less than the median.

The relative order of mean and median depends on the value of X

Answer: The volume of a sphere is given by the formula V = (4/3)Ï€r^3, where r is the radius of the sphere. Given that the radii of the 11 balls are X cm, X+1 cm, Ã‰, X+10 cm, we can write their volumes as:

V1 = (4/3)Ï€X^3

V2 = (4/3)Ï€(X+1)^3

...

V11 = (4/3)Ï€(X+10)^3

To determine which of the following statements is true about the volumes of these balls, we need to compare them.

A. V1 < V11

B. V5 > V6

C. V2 - V1 = V10 - V9

A. To compare V1 and V11, we can take the ratio of their volumes:

V1/V11 = (4/3)Ï€X^3 / (4/3)Ï€(X+10)^3 = X^3 / (X+10)^3

To simplify this expression, we can divide both the numerator and denominator by X^3:

V1/V11 = 1 / (1+10/X)^3

Since X is positive, 10/X is less than 1, so 1+10/X is greater than 1. Therefore, (1+10/X)^3 is greater than 1, which means that V1/V11 is less than 1. This implies that V1 is less than V11, so statement A is true.

B. To compare V5 and V6, we can take the ratio of their volumes:

V5/V6 = (4/3)Ï€(X+4)^3 / (4/3)Ï€(X+5)^3 = (X+4)^3 / (X+5)^3

To simplify this expression, we can expand the cube of a binomial:

V5/V6 = (X^3 + 12X^2 + 48X + 64) / (X^3 + 15X^2 + 75X + 125)

Since X is positive, we can ignore the denominators and compare the numerators:

V5 = X^3 + 12X^2 + 48X + 64

V6 = X^3 + 15X^2 + 75X + 125

To determine whether V5 is greater than V6 or not, we can subtract V6 from V5:

V5 - V6 = -3X^2 - 27X - 61

This expression is negative for all positive values of X, so V5 is always less than V6. Therefore, statement B is false.

C. To compare V2 - V1 and V10 - V9, we can subtract V1 from V2 and V9 from V10:

V2 - V1 = (4/3)Ï€[(X+1)^3 - X^3] = (4/3)Ï€(3X^2 + 3X + 1)

V10 - V9 = (4/3)Ï€[(X+10)^3 - (X+9)^3] = (4/3)Ï€(33X^2 + 297X + 811)

To determine whether V2 - V1 is equal to V10 - V9 or not, we can subtract the latter from the former:

(V2 - V1) - (V10 - V9) = (4/3)Ï€(-30X^2 - 294X - 810)

This expression is negative for all positive values of X, so V2 - V1

9. A seagull flying 30 metres above the ocean spots a fish swimming 6 metres below the surface. The seagull folds its wings and falls to catch the fish. What is the velocity of the seagull when it hits the surface of the water? Assume no air resistance.

(i) 27.5m/s

(ii) 24.25 m/s

(iii)11 m/s

(iv)588 m/s

Answer : We can solve this problem using the principles of kinematics and the conservation of energy.

When the seagull is at a height of 30 meters, it has potential energy given by:

PE = mgh

where m is the mass of the seagull, g is the acceleration due to gravity (9.8 m/s^2), and h is the height above the ocean (30 meters).

When the seagull reaches the surface of the water, it has kinetic energy given by:

KE = 1/2 mv^2

where v is the velocity of the seagull just before it hits the water.

Assuming no air resistance, the potential energy at the beginning is converted to kinetic energy at the end, so we can equate the two equations:

PE = KE

mgh = 1/2 mv^2

Solving for v, we get:

v = sqrt(2gh)

where h = 30 + 6 = 36 meters (taking into account the depth of the fish).

Plugging in the values, we get:

v = sqrt(2 x 9.8 x 36) = 24.4 m/s

Therefore, the velocity of the seagull just before it hits the surface of the water is approximately 24.4 m/s.

10. A bacterium is swimming in an isotonic medium. I now add NaCl to a concentration of 100 mM. Which of the following is true?

(i) The osmotic pressure points inward but the water flows outward

(ii) The osmotic pressure points outward so the water flows outward

(iii) The osmotic pressure points inward so the water flows inward

(iv) The osmotic pressure points outward but the water flows inward

Answer : When a bacterium is swimming in an isotonic medium, the solute concentration inside the bacterium is equal to the solute concentration outside the bacterium. Therefore, the addition of NaCl to a concentration of 100 mM may have different effects on the bacterium depending on its initial solute concentration and the type of bacterium.

If the bacterium's internal solute concentration is already 100 mM, then adding NaCl to the surrounding medium would not have any net effect on the bacterium as there would be no concentration gradient for NaCl to diffuse across the membrane.

If the bacterium's internal solute concentration is lower than 100 mM, then adding NaCl to the surrounding medium would create a concentration gradient for NaCl to diffuse into the bacterium. This could cause water to flow out of the bacterium to balance the concentration gradient, resulting in the bacterium shrinking or even dying if the osmotic stress is severe enough.

If the bacterium's internal solute concentration is higher than 100 mM, then adding NaCl to the surrounding medium would create a concentration gradient for NaCl to diffuse out of the bacterium. This could cause water to flow into the bacterium to balance the concentration gradient, resulting in the bacterium swelling or even bursting if the osmotic stress is severe enough.

Therefore, the true effect of adding NaCl to a concentration of 100 mM on a bacterium swimming in an isotonic medium depends on the initial solute concentration of the bacterium and its ability to regulate its internal solute concentration.

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