JEE Physics Units and Dimensions MCQs Set 04

Answer: (b) kilogram

Explanation:
1. Impulse, energy and density are physical quantities.
2. Kilogram is the name of a unit (the SI unit of mass), not a quantity.

Teacher's Note:
1. A quantity is something measured; a unit is the standard used to measure it.
2. Mass is the quantity, kilogram the unit.

Answer: (a) N m\(^2\) kg\(^{-2}\)

Explanation:
1. From \( F = \frac{Gm_1m_2}{r^2} \), \( G = \frac{Fr^2}{m_1m_2} \).
2. Its unit is \( \frac{\text{N m}^2}{\text{kg}^2} = \text{N m}^2\text{kg}^{-2} \).

Teacher's Note:
1. Rearrange the law and substitute SI units.
2. The r² goes in the numerator and both masses in the denominator.

Answer: (c) joule m\(^{-2}\)

Explanation:
1. Surface tension \( S = \frac{F}{l} = \frac{\text{energy}}{\text{area}} \).
2. So its unit is N m\(^{-1}\) = J m\(^{-2}\).

Teacher's Note:
1. Force per unit length equals energy per unit area.
2. Check: N/m = (N m)/m².

Answer: (c) 0

Explanation:
1. The dimensions of angular frequency are \( [T^{-1}] \), with \( L^0 \).
2. So the power of length is zero.

Teacher's Note:
1. Radian is dimensionless.
2. Only time appears in the dimension of angular frequency.

Answer: (a) \( M^{-1}L^3T^{-2} \)

Explanation:
1. \( G = \frac{Fr^2}{m^2} = \frac{(MLT^{-2})L^2}{M^2} = M^{-1}L^3T^{-2} \).

Teacher's Note:
1. Dimensions of force are MLT⁻².
2. Two masses in the denominator give M⁻¹.

Answer: (b) \( M^0L^0T^0A^{-1} \)

Explanation:
1. \( L/R \) has the dimension of time, and \( CV \) is a charge, \( [AT] \).
2. So \( \frac{L}{RCV} = \frac{T}{AT} = A^{-1} \).

Teacher's Note:
1. The L/R time constant is a standard result.
2. Charge has dimensions A T, which removes the time.

Answer: (b) \( M^0L^0T^{-1} \)

Explanation:
1. The exponent \( at + 3 \) must be dimensionless, and 3 is a pure number.
2. So \( at \) is dimensionless and a has the dimensions \( T^{-1} \).

Teacher's Note:
1. Only dimensionless numbers can appear as exponents.
2. Terms added together must have the same dimensions.

Answer: (c) stress

Explanation:
1. Stress \( = \frac{\text{force}}{\text{area}} = \frac{MLT^{-2}}{L^2} = ML^{-1}T^{-2} \).
2. Torque is \( ML^2T^{-2} \), surface tension is \( MT^{-2} \) and viscosity is \( ML^{-1}T^{-1} \).

Teacher's Note:
1. Pressure, stress and modulus of elasticity share the same dimensions.
2. Compare each option's formula with MLT⁻²/L².

Answer: (a) \( \frac{1}{x^2} \)

Explanation:
1. 1 m \( = \frac1x \) new units, so \( 1\ \text{m}^2 = \left(\frac1x\right)^2 \) new units\(^2\).

Teacher's Note:
1. A larger unit gives a smaller numerical value.
2. Areas scale as the square of the length conversion.

Answer: (d) \( 2\times10^3 \) kg/m\(^3\)

Explanation:
1. 1 g/cm\(^3\) = \( \frac{10^{-3}\text{ kg}}{10^{-6}\text{ m}^3} = 10^3 \) kg/m\(^3\).
2. So 2 g/cm\(^3\) = \( 2\times10^3 \) kg/m\(^3\).

Teacher's Note:
1. Water has density 1 g/cm³ = 1000 kg/m³.
2. Multiply the CGS value by 1000 to get SI.

Answer: (b) \( \frac{v^2}{rg} \)

Explanation:
1. \( v^2 = L^2T^{-2} \) and \( rg = L\cdot LT^{-2} = L^2T^{-2} \).
2. So \( \frac{v^2}{rg} \) is dimensionless.

Teacher's Note:
1. This is the ratio that appears in banking of roads.
2. Check the powers of L and T separately.

Answer: (d) A dimensionally correct equation may still be incorrect

Explanation:
1. Dimensional correctness is necessary but not sufficient. For example \( s = ut + at^2 \) is dimensionally consistent, yet the correct formula is \( s = ut + \tfrac12at^2 \).
2. A numerical factor such as 1/2 or 2π has no dimensions, so dimensional analysis cannot detect or fix it.
3. A dimensionally incorrect equation is always wrong.

Teacher's Note:
1. Dimensional consistency can only reject an equation, never prove it.
2. Numerical constants and trigonometric or exponential forms are beyond its reach.

Answer: (a) \( 6.67\times10^{-8} \)

Explanation:
1. 1 N = \( 10^5 \) dyne, 1 m\(^2\) = \( 10^4 \) cm\(^2\), 1 kg\(^{-2}\) = \( 10^{-6} \) g\(^{-2}\).
2. Factor \( = 10^{5+4-6} = 10^3 \), so \( G = 6.67\times10^{-11}\times10^3 = 6.67\times10^{-8} \) in CGS.

Teacher's Note:
1. Convert each unit separately and multiply the factors.
2. In CGS, G = 6.67 × 10⁻⁸ dyne cm² g⁻².

Answer: (c) 13.6 g cm\(^{-3}\)

Explanation:
1. 1 kg m\(^{-3}\) \( = \frac{1000\text{ g}}{10^6\text{ cm}^3} = 10^{-3} \) g cm\(^{-3}\).
2. So \( 13600\times10^{-3} = 13.6 \) g cm\(^{-3}\).

Teacher's Note:
1. Divide the SI value of density by 1000 to get g/cm³.
2. Mercury is 13.6 times denser than water.

Answer: (c) \( 10^5 \) N/m\(^2\)

Explanation:
1. 1 dyne = \( 10^{-5} \) N and 1 cm\(^2\) = \( 10^{-4} \) m\(^2\).
2. So 1 dyne/cm\(^2\) = \( 10^{-1} \) N/m\(^2\), and \( 10^6 \) dyne/cm\(^2\) = \( 10^5 \) N/m\(^2\).

Teacher's Note:
1. 1 N/m² = 10 dyne/cm².
2. Move the decimal point carefully when converting.

Answer: (a) \( 3.6\times10^3 \) J

Explanation:
1. 1 W h = 1 J/s \( \times \) 3600 s = \( 3.6\times10^3 \) J.

Teacher's Note:
1. A watt-hour is a unit of energy, not power.
2. 1 kWh = 3.6 × 10⁶ J.

Answer: (d) \( \pi \) cm

Explanation:
1. \( \theta = 1.8^\circ = 1.8\times\frac{\pi}{180} = \frac{\pi}{100} \) rad.
2. For small angles \( h = \theta d = \frac{\pi}{100}\times100 \text{ cm} = \pi \) cm.

Teacher's Note:
1. Small angles in radian times distance give the arc length or height.
2. Convert degrees to radian first.

Answer: (b) 3350 km

Explanation:
1. \( \theta = 0.50^\circ = 0.50\times\frac{\pi}{180} = 8.73\times10^{-3} \) rad.
2. \( D = \theta r = 8.73\times10^{-3}\times3.84\times10^5 \approx 3350 \) km.

Teacher's Note:
1. Diameter = angle (radian) × distance for small angles.
2. The actual diameter of the Moon is about 3474 km.

Answer: (b) 9.95

Explanation:
1. \( \sqrt{99} = 10\sqrt{1 - 0.01} = 10(1 - 0.01)^{1/2} \).
2. \( \approx 10\left(1 - \frac{0.01}{2}\right) = 10(0.995) = 9.95 \).

Teacher's Note:
1. Write the number as a power of ten times (1 + x).
2. The exact value is 9.9499.

Answer: (c) 0.99

Explanation:
1. \( \frac{1}{1.01} = (1 + 0.01)^{-1} \approx 1 - 0.01 = 0.99 \).

Teacher's Note:
1. A negative power gives 1 − nx with n = 1.
2. The exact value is 0.990099.

Answer: (a) 0.14 and 0.087

Explanation:
1. \( 8^\circ = 8\times\frac{\pi}{180} = 0.1396 \) rad, so \( \sin8^\circ \approx 0.14 \).
2. \( 5^\circ = 0.0873 \) rad, so \( \tan5^\circ \approx 0.087 \).

Teacher's Note:
1. For small angles in radian, sin θ ≈ tan θ ≈ θ.
2. Never put the angle in degrees directly.

Answer: (d) \( \frac{1}{16} \) and \( \frac14 \)

Explanation:
1. Area \( \propto r^2 \): \( \frac{r^2}{16r^2} = \frac1{16} \).
2. Perimeter \( \propto r \): \( \frac{r}{4r} = \frac14 \).

Teacher's Note:
1. Areas scale as the square of the linear size, and perimeters scale linearly.
2. Do not square the perimeter ratio.

Answer: (b) \( \frac{148\pi}{3} \) cm\(^3\)

Explanation:
1. \( V = \frac43\pi(R^3 - r^3) = \frac43\pi(64 - 27) = \frac{148\pi}{3} \) cm\(^3\).

Teacher's Note:
1. Volume of material = outer volume minus inner volume.
2. Subtract the cubes, not the cube of the difference.

Answer: (d) \( -\frac12 \)

Explanation:
1. \( \cos120^\circ = \cos(180^\circ - 60^\circ) = -\cos60^\circ = -\frac12 \).

Teacher's Note:
1. Cosine is negative in the second quadrant.
2. Use the reference angle 60°.