Question 41:
easyAssertion (A): When we change the unit of measurement of a quantity, its numerical value changes.
Reason (R): Smaller the unit of measurement smaller is its numerical value.
The physical magnitude is invariant, expressed as \(n u = \text{constant}\). Therefore, numerical value is inversely proportional to the unit size. A smaller unit leads to a larger numerical value, making R false.
Question 42:
easyThe vernier scale of a callipers is divided into 20 divisions which coincide with 18 main scale divisions. Each main scale division is 0.2 mm. The least count of the instrument is
Least count (LC) is given by \(text{LC} = 1text{ MSD} - 1text{ VSD}\). Since 20 VSD = 18 MSD, we have \(1text{ VSD} = 0.9text{ MSD}\). Thus, \(text{LC} = 0.1text{ MSD} = 0.1 times 0.2text{ mm} = 0.02text{ mm}\).
Question 43:
easyTwo quantities are measured as \( P = (1 \pm 0.40) \, \text{m} \), \( Q = (4 \pm 0.20) \, \text{m} \). The correct value of \( (PQ)^{1/2} \) will be
Let \( Y = (PQ)^{1/2} \). Its value is \( Y = (1 \times 4)^{1/2} = 2 \, \text{m} \). The relative error is \( \frac{\Delta Y}{Y} = \frac{1}{2} \left(\frac{\Delta P}{P} + \frac{\Delta Q}{Q}\right) = \frac{1}{2}\left(\frac{0.40}{1} + \frac{0.20}{4}\right) = 0.225 \), giving \( \Delta Y = 0.45 \, \text{m} \). Thus, \( Y = (2 \pm 0.45) \, \text{m} \).
Question 44:
easyDashrath measures the length of a wire using a meter scale with a least count of \( 1\text{ mm} \) and finds it to be \( L = 75.0\text{ cm} \). He also measures diameter of thin wire using a screw gauge with a least count of \( 0.01\text{ mm} \) and finds it to be \( d = 0.500\text{ cm} \). He uses these measurements to calculate the volume of wire. The maximum percentage error in volume of wire is nearly
Volume \( V = \frac{\pi d^2 L}{4} ⇒ \frac{\Delta V}{V} = 2 \frac{\Delta d}{d} + \frac{\Delta L}{L} \). Here, \( \Delta d = 0.001\text{ cm} \) and \( \Delta L = 0.1\text{ cm} \). Thus, \( \frac{\Delta V}{V} = 2\left(\frac{0.001}{0.500}\right) + \frac{0.1}{75} = 0.4% + 0.13% = 0.53% \).
Question 45:
easyThe dimensional formula for impulse is
Impulse is defined as Force multiplied by time, \( I = F \cdot t \). Its dimensions are \( [\text{MLT}^{-2}][\text{T}] = [\text{MLT}^{-1}] \).
Question 46:
easyAssertion (A): In SHM let \(x\) be the maximum speed, \(y\) the frequency of oscillation and \(z\) the maximum acceleration, then \(\frac{xy}{z}\) is a constant quantity.
Reason (R): This is because \(\frac{xy}{z}\) becomes a dimensionless quantity
For SHM, \(x=A\omega\), \(y=\frac{\omega}{2\pi}\), \(z=A\omega^2\). Thus, \(\frac{xy}{z} = \frac{(A\omega)(\omega/(2\pi))}{(A\omega^2)} = \frac{1}{2\pi}\), which is a constant. So (A) is true. The dimensions are \([x]=LT^{-1}\), \([y]=T^{-1}\), \([z]=LT^{-2}\), making \([xy/z]=1\), dimensionless. So (R) is true. However, being dimensionless does not explain why it's a constant.
Question 47:
easyThe dimensions of \((\mu_0\epsilon_0)^{-1/2}\) are:
[2012 Mains]
The speed of light \(c = 1/\sqrt{\mu_0\epsilon_0}\). Therefore, \((\mu_0\epsilon_0)^{-1/2} = c\). The dimensions of speed are \([LT^{-1}]\).
Question 48:
easyThe dimension of \(\frac{1}{2}\epsilon_0 E^2\), where \(\epsilon_0\) is permittivity of free space and \(E\) is electric field, is:
[2010 Pre]
The expression \(\frac{1}{2}\epsilon_0 E^2\) represents electric energy density, which is Energy per unit Volume. \([\text{Energy density}] = [\text{Energy}]/[\text{Volume}] = (ML^2T^{-2})/L^3 = ML^{-1}T^{-2}\).
Question 49:
easyThe dimensions of universal gravitational constant are:
[2004]
From \(F = G \frac{m_1 m_2}{r^2}\), we get \(G = \frac{Fr^2}{m_1 m_2}\). So, \([G] = \frac{(MLT^{-2})(L^2)}{M^2} = M^{-1}L^3T^{-2}\).
Question 50:
easyIf \(x = at + bt^2\), where \(x\) is the distance travelled by the body in kilometres while \(t\) is the time in seconds, then the units of \(b\) is:
[1989]
By homogeneity, \([bt^2] = [x]\). So \([b] = [x]/[t^2] = km/s^2\).