Units and Measurements — JEE Main practice

1. In the term

   $$h+Bt,$$

   both quantities being added must have the same dimensions.

   Since $h$ represents height,

   $$[h]=L.$$

   Therefore,

   $$[B][t]=L.$$

   As

   $$[t]=T,$$

   we obtain

   $$[B]=LT^{-1}=[M^0LT^{-1}].$$

2. In the term

   $$P+\frac{A}{Bt^2},$$

   the quantity

   $$\frac{A}{Bt^2}$$

   must have the same dimensions as pressure.

3. Pressure is force per unit area, so

   $$[P]=\frac{MLT^{-2}}{L^2}=ML^{-1}T^{-2}.$$

4. Since

   $$[B]=LT^{-1},$$

   we have

   $$[Bt^2]=(LT^{-1})(T^2)=LT.$$

5. Therefore,

   $$\left[\frac{A}{Bt^2}\right]=ML^{-1}T^{-2},$$

   which gives

   $$[A]=(ML^{-1}T^{-2})(LT)=MT^{-1}.$$

   Hence,

   $$[A]=[ML^0T^{-1}].$$

6. Therefore, the required dimensions are

   $$[A]=[ML^0T^{-1}],\qquad [B]=[M^0LT^{-1}].$$

   Hence, the correct answer is Option B.

1. When two springs are connected in parallel, both experience the same extension.

   Therefore,

   $$F_1=k_1x,\qquad F_2=k_2x.$$

2. The total restoring force is the sum of the individual forces:

   $$F=F_1+F_2.$$

   Hence,

   $$k_{\mathrm{eq}}x=k_1x+k_2x,$$

   giving

   $$k_{\mathrm{eq}}=k_1+k_2.$$

3. Substituting the given values:

   $$k_{\mathrm{eq}}=10+20=30\ \mathrm{N/m}.$$

4. For the sum of measured quantities, the absolute errors are added:

   $$\Delta k_{\mathrm{eq}} = \Delta k_1+\Delta k_2 = 0.2+0.3 0.5\ \mathrm{N/m}.$$

5. The percentage error is:

   $$\frac{\Delta k_{\mathrm{eq}}}{k_{\mathrm{eq}}}\times100 = \frac{0.5}{30}\times100 1.666\ldots$$

6. Rounding to two decimal places:

   $$1.666\ldots$$

7. Therefore, the equivalent spring constant is

   $$(30\pm0.5),\mathrm{N/m},$$

   which corresponds to a percentage error of

Hence, the correct answer is Option B.

1. For addition and subtraction, the final result should be reported with the same number of decimal places as the measurement having the fewest decimal places.

2. The given measurements are:

   • $52.01\ \mathrm{m}$ → 2 decimal places
   • $153.2\ \mathrm{m}$ → 1 decimal place
   • $0.123\ \mathrm{m}$ → 3 decimal places

3. Therefore, the final answer must be reported to 1 decimal place, since $153.2\ \mathrm{m}$ has the least number of decimal places.

4. Add the values:

   $$52.01+153.2+0.123=205.333\ \mathrm{m}$$

5. Round the result to one decimal place:

   $$205.333\ \mathrm{m}\approx205.3\ \mathrm{m}$$

6. Hence, the required sum is:

   $$\boxed{205.3\ \mathrm{m}}$$

   Therefore, the correct answer is Option B.