In a measurement, it is asked to find modulus of elasticity per unit torque applied on the system. The measured quantity has dimension of [MaLbTc]$\left[\mathrm{M}^{\mathrm{a}}\mathrm{L}^{\mathrm{b}}\mathrm{T}^{\mathrm{c}}\right]$ . If b = 3$b = 3$ , the value of c$c$ is

Related Formula

Core Logic

Since the zero of the vernier scale is to the right of the main scale zero, there is a positive zero error. We must subtract this zero error from the observed reading.

Step 1: Calculate Zero Error

Zero error = +(4 × 0.1 ~mm) = +0.4 ~mm$= +(4 \times 0.1 \mathrm{~mm}) = +0.4 \mathrm{~mm}$

Step 2: Calculate Final Reading

Observed Reading = 15 ~mm + (5 × 0.1 ~mm) = 15.5 ~mm$= 15 \mathrm{~mm} + (5 \times 0.1 \mathrm{~mm}) = 15.5 \mathrm{~mm}$

True Reading = Observed Reading - Zero Error$= \text{Observed Reading} - \text{Zero Error}$ = 15.5 ~mm - 0.4 ~mm$= 15.5 \mathrm{~mm} - 0.4 \mathrm{~mm}$ = 15.1 ~mm$= 15.1 \mathrm{~mm}$

Pattern Recognition

Vernier zero right of main zero arrow$\rightarrow$ positive error arrow$\rightarrow$ subtract it. Vernier zero left of main zero arrow$\rightarrow$ negative error arrow$\rightarrow$ add it.

Chapter Mix

Class 11 Physics: Units and Measurements

Core Logic

The given readings are 1.24 mm$1.24 \text{ mm}$, 1.25 mm$1.25 \text{ mm}$, 1.23 mm$1.23 \text{ mm}$, 1.21 mm$1.21 \text{ mm}$. Each of these readings is recorded up to two decimal places in millimeters.

Step 1: Determine Least Count

The least count of an instrument represents the smallest value that it can measure accurately, which is typically indicated by the decimal precision of the recorded data. Since the smallest variation possible between readings written to two decimal places is 0.01 mm$0.01 \text{ mm}$ (e.g., the difference between 1.24$1.24$ and 1.23$1.23$), the least count must be 0.01 mm$0.01 \text{ mm}$.

Pattern Recognition

Look at the decimal places. If a measurement is X.YZ$X.YZ$, the least count is generally the place value of the last significant digit, i.e., 0.01$0.01$.

Chapter Mix

Class 11 Physics: Units and Measurements

Related Formula

(Note: The question states T = 2π √(k/m)$T = 2\pi \sqrt{k/m}$, which is dimensionally incorrect. Assuming it meant T = 2π √(m/k)$T = 2\pi \sqrt{m/k}$ or just working with the error propagation mathematically.) Regardless, rearranging for k$k$ gives k ∝ (m)/(T²)$k \propto \frac{m}{T^2}$ or k ∝ mT²$k \propto mT^2$. Using standard error analysis logic for multiplication/division:

Step 1: Extract Given Data

m = 10 g = 10 × 10⁻³ kg$m = 10 \text{ g} = 10 \times 10^{-3} \text{ kg}$ Δ m = 10 mg = 10 × 10⁻³ g$\Delta m = 10 \text{ mg} = 10 \times 10^{-3} \text{ g}$ Total time for N=50$N=50$ oscillations is t = 60 s$t = 60 \text{ s}$. Resolution Δ t = 2 s$\Delta t = 2 \text{ s}$. Time period T = (t)/(N) = (60)/(50) = 1.2 s$T = \frac{t}{N} = \frac{60}{50} = 1.2 \text{ s}$. Error in Time period Δ T = (Δ t)/(N) = (2)/(50) s$\Delta T = \frac{\Delta t}{N} = \frac{2}{50} \text{ s}$.

Step 2: Calculating Fractional Errors

Wait, the solution uses (2)/(50 × 1.2)$\frac{2}{50 \times 1.2}$. Let's apply it properly:

Step 3: Calculating Percentage Error in K

Percentage error:

Pattern Recognition

For total time t = NT$t = NT$, the relative error (Δ T)/(T)$\frac{\Delta T}{T}$ is exactly equal to (Δ t)/(t)$\frac{\Delta t}{t}$. Don't overcomplicate by dividing by N$N$ separately for both, it cancels out.

Chapter Mix

Class 11 Physics: Units and Measurements Class 11 Physics: Oscillations

Core Logic

(A) Coefficient of viscosity (η$\eta$):

(B) Surface tension (S$S$):

(C) Pressure (P$P$):

(D) Surface energy (E$E$): Energy has the dimensions of work.

Step 1: Final Match

A arrow$\rightarrow$ IV B arrow$\rightarrow$ III C arrow$\rightarrow$ I D arrow$\rightarrow$ II

Chapter Mix

Class 11 Physics: Units and Measurements Class 11 Physics: Mechanical Properties of Fluids

Related Formula

Let momentum be P$P$ ([P] = [M L T⁻¹]$[P] = [M L T^{-1}]$). We assume:

Core Logic

Let's find the dimensional formulas of the individual variables:

• Charge (q$q$):

[q] = [A T]$[q] = [A T]$

• Current (I$I$):

[I] = [A]$[I] = [A]$

• Permeability of vacuum (μ₀$\mu_0$):

From Biot-Savart law or force between parallel wires: F = (μ₀ I² L)/(2π d)$F = \frac{\mu_0 I^2 L}{2\pi d}$:

Step 1: Apply Dimensional Homogeneity

Substitute these into our assumed dimensional equation:

Comparing exponents on both sides:

• For [M]$[M]$:

y = 1$y = 1$

• For [L]$[L]$:

• For [T]$[T]$:

• For [A]$[A]$:

Thus, x = 1, y = 1, z = 1$x = 1, y = 1, z = 1$.

Therefore, the required quantity is:

Pattern Recognition

Sees: Permeability, charge, and current linked to momentum. Trap: Deriving the dimensions of μ₀$\mu_0$ using complex magnetic formulas. Remember [μ₀] = [Force]/[Current]²$[\mu_0] = [\text{Force}]/[\text{Current}]^2$ is the quickest way to get its dimensions. Shortcut: Since [q] = AT$[q] = AT$ and [I] = A$[I] = A$, [q μ₀ I] = [A T] [M L T⁻² A⁻²] [A] = [M L T⁻¹]$[q \mu_0 I] = [A T] [M L T^{-2} A^{-2}] [A] = [M L T^{-1}]$, which is exactly the dimensions of momentum. Hence, option (2) is correct.

Chapter Mix

Class 11 Physics: Units and Measurements Class 12 Physics: Moving Charges and Magnetism