A: Force is not a base SI quantity. It is derived using mass m and acceleration a in the formula F = m a .

B: newton is a SI unit and not a SI quantity.

C: Second is a base SI unit not a SI quantity.

D: time is a base SI quantity .

You need to convert 50 km h^-1 to m s^-1

50 km h^-1 = 50 000 m ÷ 3600 s = 14 m s^-1

Assume the mass of a car is 1500 kg

KE = 1 /2 m v²

KE = 1 / 2 x 1500 x (14)² = 147 000 J is rounded 1.5 x 10⁵ J

Answer B is correct.

The resistance of 100 cm wire = 10 Ω → R branch RST = 10 Ω

The resistance of 50 cm wire = 5 Ω → R branch RT = 5 Ω

The resistance of 25 cm wire = 2.5 Ω → R branch XR = 2.5 Ω and R branch TY = 2.5 Ω

The circuit can be considered to be built up as shown below:

R between R and T : ` 1/R = 1/5 + 1/10`

→ ` 1/R = 2/10 + 1/10 = 3/10`

R between R and T `= 10/3 = 3.3 Ω`

R between X Y = 2.5 Ω + 3.3 Ω + 2.5 Ω = 8.3 Ω

Answer C is correct.

A double slit experiment is shown below.

A maximum at X_n on the screen will be formed when the difference in distance travelled by light from slit 1 and travelled from slit 2 = n λ (= n wavelengths).

The 2 waves arriving from slit 1 and slit 2 are in phase.

This distance (n λ) is shown in the diagram below in the small triangle close to the slits.

In this triangle the following relation is applicable:

`sin θ = n λ/a`
(1) (a is the distance between the centres of the slits)

In the triangle of the central axis the following relation is applicable:

`tan θ = X_n/D` (2)

If θ is small (smaller than 15 ^o) then sin θ is nearly equal to tan θ.

(sin 15^o = 0.26 and tan 15^o = 0.27)

Combining (1) and (2) →

`n λ/a = X_n/D` → `X_n = n λ D/a` (3)

(this relation (3) is not exact, but an acceptable approximation)

From relation (3) can be concluded:

The distance between fringes is proportional to

1. the wavelength λ of the light and
2. the distance D between the slits and the screen.

However, the distance between the fringes is inversely proportional to the separation a of the slits.

Answer A is wrong. The distance between the light source and the slits does not affect the distance between the fringes.

The correct answer is B: increasing the distance between the slits and the screen will increase the distance between the fringes.

Answer C is wrong. If the distance between the slits increases the distance between the fringes will decrease.

Answer D is wrong. The wavelength is smaller at a higher frequency and this will decrease the distance between the fringes.