JEE Physics Motion In A Plane MCQs Set 19

Answer: (c) 272 m

Explanation:
1. Maximum height (thrown vertically) \( H=\frac{u^2}{2g}=136\ \text{m} \).
2. Maximum range (at 45°) \( R=\frac{u^2}{g}=2H=272\ \text{m} \).

Teacher's Note:
1. The man throws with the same speed u in both cases.
2. Maximum range is twice the maximum height.

Answer: (a) 40 m/s

Explanation:
1. \( v=\frac{dx}{dt}=8t=8\times5=40\ \text{m/s} \).

Teacher's Note:
1. Velocity is the first derivative of position with respect to time.
2. Units follow from x in metres and t in seconds.

Answer: (a) 120 m/s²

Explanation:
1. Speed just before impact \( v_1=\sqrt{2\times10\times9.8}=14\ \text{m/s} \) (downward).
2. Speed just after impact \( v_2=\sqrt{2\times10\times5}=10\ \text{m/s} \) (upward).
3. Change of velocity \( =10-(-14)=24\ \text{m/s} \), so \( a=\frac{24}{0.2}=120\ \text{m/s}^2 \).

Teacher's Note:
1. The velocity reverses direction, so the change is the sum of the two speeds.
2. Use v² = 2gh for each part of the motion.

Answer: (a) A-II, B-IV, C-III, D-I

Explanation:
1. The slope of an x-t graph gives the velocity.
2. A: slope increases linearly (parabola), so v increases linearly: II.
3. B: x falls (negative slope) with decreasing magnitude, so v starts at -v0 and tends to zero: IV.
4. C: positive constant slope then equal negative constant slope: III.
5. D: constant slope, so constant v: I.

Teacher's Note:
1. Read the gradient of each x-t graph and sketch the v-t graph.
2. Constant velocity means a straight x-t line.

Answer: (d) 3.75 km/h

Explanation:
1. Times: \( \frac43+\frac45=\frac{32}{15}\ \text{h} \).
2. \( v_{avg}=\frac{8}{32/15}=3.75\ \text{km/h} \).

Teacher's Note:
1. Equal distances at different speeds: the average is the harmonic mean, not the arithmetic mean.
2. Average speed = total distance / total time.

Answer: (a) 3 km/h

Explanation:
1. Time to cross \( t=\frac{1\ \text{km}}{4\ \text{km/h}}=0.25\ \text{h} \).
2. Drift \( =v_rt\Rightarrow0.75=v_r\times0.25 \), so \( v_r=3\ \text{km/h} \).

Teacher's Note:
1. Heading perpendicular to the flow, crossing time depends only on the swimmer's own speed.
2. The downstream drift is caused only by the river current.

Answer: (c) \( \frac{3v_1v_2v_3}{v_1v_2+v_2v_3+v_3v_1} \)

Explanation:
1. Since AD = 3 AB and AB = BC, all three segments have the same length s.
2. Total time \( =s\left(\frac1{v_1}+\frac1{v_2}+\frac1{v_3}\right) \).
3. \( v_{avg}=\frac{3s}{s\left(\frac1{v_1}+\frac1{v_2}+\frac1{v_3}\right)}=\frac{3v_1v_2v_3}{v_1v_2+v_2v_3+v_3v_1} \).

Teacher's Note:
1. Equal distances at different speeds give the harmonic mean.
2. Check with v1 = v2 = v3 = v: the result is v.

Answer: (b) 15 m/s

Explanation:
1. Energy conservation: \( v^2=u^2+2gh=25+2\times10\times10=225 \).
2. \( v=15\ \text{m/s} \).

Teacher's Note:
1. The horizontal speed stays 5 m/s; the vertical speed is √(2gh) = 14.1 m/s.
2. Combining them gives √(25 + 200) = 15 m/s.

Answer: (a) 1

Explanation:
1. Perpendicular vectors have zero dot product: \( 2a-3b+4=0 \).
2. With \( 3a+2b=7 \): multiply the first by 2 and the second by 3: \( 4a-6b=-8,\ 9a+6b=21 \), so \( 13a=13\Rightarrow a=1,\ b=2 \).
3. \( \frac ab=\frac12=\frac x2 \), so x = 1.

Teacher's Note:
1. Set up the dot product and solve the two linear equations.
2. Check: 2(1) - 3(2) + 4 = 0.

Answer: (a) 4

Explanation:
1. \( \vec P\times\vec Q=\hat i(2.5\sqrt3-2\sqrt3)-\hat j(7.5-8)+\hat k(3\sqrt3-4\sqrt3)=\frac{\sqrt3}{2}\hat i+\frac12\hat j-\sqrt3\hat k \).
2. Magnitude \( =\sqrt{\frac34+\frac14+3}=2 \).
3. Unit vector \( =\frac14\left(\sqrt3\hat i+\hat j-2\sqrt3\hat k\right) \), so x = 4.

Teacher's Note:
1. Use the determinant form of the cross product.
2. Divide by the magnitude to get the unit vector.

Answer: (a) -30 J

Explanation:
1. Displacement \( \vec d=(5-2)\hat i+(-2-3)\hat j+(1+4)\hat k=3\hat i-5\hat j+5\hat k \).
2. \( W=\vec F\cdot\vec d=15-10-35=-30\ \text{J} \).

Teacher's Note:
1. Work is the dot product of force and displacement.
2. The z-component contributes -7 x 5 = -35 J.

Answer: (a) 13 m/s

Explanation:
1. Friction supplies the centripetal force: \( \mu mg=\frac{mv^2}{r} \).
2. \( v=\sqrt{\mu rg}=\sqrt{0.34\times50\times10}=\sqrt{170}\approx13\ \text{m/s} \).

Teacher's Note:
1. The maximum speed does not depend on the mass of the car.
2. Maximum static friction = μ N = μ m g.

Answer: (a) 40

Explanation:
1. \( \frac{dv}{dt}=\frac{v^2}{R}\Rightarrow\frac{dv}{v^2}=\frac{dt}{R}\Rightarrow\frac1u-\frac1v=\frac tR \).
2. Also \( v\frac{dv}{ds}=\frac{v^2}R\Rightarrow v=ue^{s/R} \). After a quarter circle \( s=\frac{\pi R}{2} \): \( v=ue^{\pi/2} \).
3. \( t=R\left(\frac1u-\frac1v\right)=\frac Ru\left(1-e^{-\pi/2}\right)=\frac{600}{15}\left(1-e^{-\pi/2}\right) \), so t = 40 (u = 54 km/h = 15 m/s).

Teacher's Note:
1. Use v dv/ds to relate speed and arc length.
2. Convert 54 km/h into 15 m/s.

Answer: (d) \( 10\sqrt2 \)

Explanation:
1. In 3 s the body has completed three quarters of a revolution, i.e. the angle turned is 270°.
2. The start and end points are at the ends of two perpendicular radii: \( d=\sqrt{10^2+10^2}=10\sqrt2\ \text{m} \).

Teacher's Note:
1. Displacement is the straight line distance between the start and the end points.
2. The chord for a 90° (or 270°) angle is r√2.