1. Dimensional analysis

Verify the correctness of the relation: s = ut + ½ at^2 by the method of dimensions

Dimensions:

• displacement (s) = L
• velocity (u) = LT−1
• acceleration (a) = LT−2
• time (t) = T

Dimensional analysis: [L] = [LT−1][T] + ½ [LT−2][T]2 => [L] = [L][T]−1 + ½ [L] [T]−2 => [L] = [L] + [L] => [L] = [L]

Since both sides have the same dimensions, the relation s = ut + 1/2at2 is dimensionally correct.

Show P = mv^2/R is dimensionally correct with P standing for centripetal force, m is the mass, v the tangential(linear) velocity and R the radius of curvature of the path.

Centripetal force (P): Force is the product of mass (m) and acceleration (a) => [P] = [M][LT−2] Tangential velocity (v): [V] = [LT-1] Radius of curvature (R): [R] = [L] Dimensional analysis: [M][LT−2]=[M][LT−1]2/[L] Simplifying the equation, [M][L]−1[T]−2=[M][L]2[T]−2[L]−1 => [L]-1=[L][L]-1 => [L]-1=[L]0 (L raise to the power zero equals to 1) => [L]-1=1 => [P] = [MLT−2] Hence the equation P = mv2/R is dimensionally correct.

By the method of dimensions only, show that the equation of motion is dimensionally correct, i.e. [M][L][T]−2=[M][L][T]−1.**

Dimensions:

• force (F) = [MLT−2]
• mass (m) = [M]
• acceleration (a) = [LT−2]

Dimensional analysis: [M][LT−2]=[M][LT]−1 Simplifying the equation, [L][T]−1=[T][L][T]−1 => [L][-1]=1L[T]-1 => [L]0=[L]0[T]0 =>1=1

Since both sides have the same dimensions, the equation of motion is dimensionally correct.

Show that the formula h = gmR2/GM holds good dimensionally. (Here, h is the height above earth’s surface where effective acceleration due to gravity is h/16, R is earth’s radius, g the acceleration due to gravity at the surface of the earth, and GM is the product of gravitational constant G and mass of the earth)

Dimensions:

• height (h) = [L]
• acceleration due to gravity (g) = [LT-2]
• radius of the Earth (R) = [L]
• gravitational constant (G) = [M−1L3T−2]
• mass of the Earth (M) = [M]

Dimensional analysis: [L]=[LT−2][L]2/[M][M] Simplifying the equation, [M]1[L]1[T]2=[M]2[L]3[T]2 =>[M]2[L]4=[M]2[L]3 =>[L]=[L] Since both sides have the same dimensions, the formula h = gmR2/GM is dimensionally correct.

Show dimensionally that the excess pressure P of a soap bubble exceeds 4S/R, where S is the surface tension, and R is the radius of the bubble.

Excess pressure (P) = [ML−1T−2] Surface tension (S) = [MT−2] Radius (R) = [L]

Dimensional analysis: [P]=4[S]/R Simplifying, [ML−1T−2]=[MT−2]/[L] => [M][L]2[T]−2= [M][T]−2 => [L]2=1 => [L] = [1]1⁄2 => [L] = [L]0 Therefore, the excess pressure P exceeds 4S/R dimensionally.

If E=hv/λ where h is Plank’s constant, v is the frequency and λ is the wavelength of a photon of light, what are the dimensions of Plank’s constant?

Energy (E) = [ML2T−2] Frequency (v) = [T−1] Wavelength (λ) = [L] Planck’s constant (h) = [ML2T−1]

Dimensional analysis: [ML2T−2]=[ML2T−1]/[L] Simplifying the above equation, [M][L]2[T]−1=[M][L]2[T]−2 Therefore, [T]=[T]0

Hence the dimensions of Plank’s constant are [ML2T−1].

If F=ma, M=kl3, and a=v2/R, show that [l]=[M1​T2L−1]​

Force (F) = [MLT−2] Mass (M) = [K][L3] Acceleration (a) = [LT−2]

Dimensional analysis: [MLT−2]=[K][L3][LT−2] Simplifying, [M][L]3[T]−2= [K][L]3[T]−2 Therefore, [L] = [M1​T2L−1]​

2. Units and Conversion of units:

Converting 2000 miles per hour to meter per second

1 mile = 1609.34 meters 1 hour = 60 minutes = 3600 seconds

So 2000 miles per hour = 2000 * 1609.34 meters / (3600 * 1) seconds = 894.08 meters per second.

Converting 20o Celsius to Fahrenheit

We know that (°C=((°F−32)5⁄9). So, (20°C=\left ( \frac59(F−32)\right ))

(20×9/5+32=F) (36+32=F) (F=68)

So 20 °C = 68 °F.

1 light-year = how many kilometers?

1 light-year = 9.46 × 10^12 km.

3. Significant figures and Uncertainities

Write 250 gm in scientific notation and mention the number of significant figures in it.

250 gm = 0.250 kg and it contains three significant figures.

Which measurement 2.345 m or 2.35 m has more accuracy and precision?

More significant digits or more decimal places do not necessarily imply greater accuracy or precision. While 2.35 m does indeed have more significant digits than 2.345 m, it does not necessarily mean it is more accurate. The number of significant figures reveals the level of uncertainty in the measurement. Usually, the last digit in a measurement carries some uncertainty. This implies that the measurement of 2.35 m, although having more decimal places, is less accurate than 2.345 m. When we claim a measurement to be 2.345 meters, the measurement falls between 2.3445 meters and 2.3455 meters. So the maximum possible error is 0.0005 meters. On the other hand, if we claim a measurement to be 2.35 meters, the measurement falls somewhere between 2.345 and 2.355 meters, which means a maximum possible error of 0.005. Hence 2.345 meters is the more accurate value between the two measurements provided.

4. Dimensional Analysis and Problem-solving

A body is thrown vertically upwards. What will be the ratio of the distance covered by the body during the last two seconds of its ascent to the distance covered by it in the first two seconds of ascent?

Let (v) be the velocity with which the body is projected upwards.
Then the distances covered in the first two seconds and last two seconds of the ascent are given by, $$s_1=u(2)-12(2)^2=2u-2g$$ $$s_2=u(t_2)-12gt^2=u(t_2-2)-g(t_