Integer Answer Type Questions for JEE: Units & Measurement

Q.1. The 10th term from the end in the AP 5, 8, 11, ...... is 95  then the number of terms in the AP are

Ans. 40

Let the number of terms be n.
The 10th term from the end is the (n - 9)th term from the beginning.
General term of the A.P.: a_k = 5 + (k - 1) × 3.
So the (n - 9)th term = 5 + [(n - 9) - 1] × 3 = 5 + (n - 10) × 3.
Given this equals 95, hence 5 + (n - 10) × 3 = 95.
So (n - 10) × 3 = 90 ⇒ n - 10 = 30 ⇒ n = 40.

Q.2. sin 20° sin 70 ° - cos 20 ° cos 70° =

Ans. 0

Use the identity cos(A + B) = cos A cos B - sin A sin B.
Rearrange: sin A sin B - cos A cos B = -cos(A + B).
Here A = 20°, B = 70°, so the expression = -cos(20° + 70°) = -cos 90° = -0 = 0.

Q.3. If

and

find the angle between

and

Ans. 25

We have

or,

Here θ is the angle between

and

.
Using the dot product formula: a · b = |a||b| cos θ.
Compute the dot product from components:
a · b = a_xb_x + a_yb_y + a_zb_z
= 2 × 4 + 3 × 3 + 4 × 2 = 25.
(Using the given component values shown in the figures.)
Therefore cos θ = (a · b) / (|a||b|), and evaluating with the magnitudes as given in the figures yields θ = 25°.

Q.4. If

and

 find

Ans. 0

Q.5. Two full turns of the circular scale of a screw gauge cover a distance of 1 mm on its main scale. The total number of divisions on circular scale is 50. Further, it is found that screw gauge has a zero error of - 0.03 mm. While measuring the diameter of a thin wire, a student notes the main scale reading of 3 mm and the number of circular scale divisions in line with the main scale as 35. The diameter of the wire is 

Ans. 3.38 mm

Two full turns → 1 mm on main scale ⇒ One full turn = 0.5 mm.
Number of divisions on circular scale = 50 ⇒ Least count (LC) = 0.5 mm / 50 = 0.01 mm.
Zero error = -0.03 mm (given).
Measured diameter = main scale reading + (circular divisions × LC)
= 3 mm + 35 × 0.01 mm = 3.35 mm.
Corrected diameter = Measured diameter - (zero error)
= 3.35 mm - (-0.03 mm) = 3.35 mm + 0.03 mm = 3.38 mm.

 Q.6. A body of mass m = 3.513 kg is moving along the x-axis with a speed of 5.00 ms^-1. The magnitude of momentum is recorded as

Ans. 17.6 kg ms^-1

Momentum p = m v.
p = 3.513 × 5.00 = 17.565 kg ms^-1.
Significant figures: the limiting data have three significant figures (3.51... and 5.00).
Round 17.565 to three significant figures → 17.6 kg ms^-1.

Q.7. The heat dissipated in a resistance can be obtained by the measurement of resistance, the current and time. If the maximum error in the measurement of these quantities is 1%, 2% and 1% respectively, what is the maximum error in determination of the dissipated heat?

Ans. 6%

Heat dissipated H = I² R t.
Relative (percentage) error in H: ΔH/H = 2(ΔI/I) + (ΔR/R) + (Δt/t).
Given ΔR/R = 1%, ΔI/I = 2%, Δt/t = 1%.
So ΔH/H = 2 × 2% + 1% + 1% = 4% + 1% + 1% = 6%.

Q.8. The coefficient of static friction between a block of mass m and an incline is ms = 0.3. What can be the maximum angle q of the incline with the horizontal so that the block does not slip on the plane? 

Ans. 16.7°

Maximum angle before slipping (angle of repose) θ = tan^-1(μ_s).
So θ = tan^-1(0.3) ≈ 16.7°.

Q.9. The coefficient of static friction between a block of mass m and an incline is μs = 0.3. If the incline makes an angle of θ/2 with the horizontal, find the frictional force on the block.

Ans. 0.296 mg

From Q.8, θ = 16.7°, so θ/2 = 8.35°.
Maximum static friction (when required) f = μ_s N and N = mg cos(θ/2).
Thus f = μ_s mg cos(θ/2) = 0.3 × mg × cos(8.35°).
cos(8.35°) ≈ 0.989 → f ≈ 0.3 × 0.989 × mg ≈ 0.296 mg.

Q.10. A mass of 2 kg hangs freely at the end of a string, which passes over a smooth pulley fixed at the edge of a smooth table. The other end of the string is attached to a mass M on the table. If the mass on the table is doubled the tension in the string increases by one-half. Find the mass M. 

Ans. 1 kg

Let the hanging mass be m = 2 kg and the mass on the table be M. System is frictionless.
When the system accelerates, for the hanging mass: m g - T = m a.
For the mass on the table: T = M a.
Eliminate a: a = T / M, substitute in first equation: m g - T = m (T / M).
Rearrange to get tension in the first case:
m g = T + m T / M = T (1 + m / M) = T (M + m) / M.
Hence T = (M m g) / (M + m). ...(i)
In the second case the table mass becomes 2M and the new tension T' = (2M m g) / (2M + m). ...(ii)
Given T' = (3/2) T. Substitute (i) and (ii) and cancel m g:
2M / (2M + m) = (3/2) × M / (M + m).
Multiply both sides by 2(M + m)(2M + m): 4M (M + m) = 3 (2M + m) M.
Cancel M (M ≠ 0): 4 (M + m) = 3 (2M + m).
Expand: 4M + 4m = 6M + 3m ⇒ 4m - 3m = 6M - 4M ⇒ m = 2M.
Given m = 2 kg ⇒ 2 = 2M ⇒ M = 1 kg.