JEE Physics Mathematical Tools MCQs Set 05

Answer: (b) \( 1+2x+\dfrac2{x^3}-\dfrac1{x^2} \)

Explanation:
1. Expand: \( y=x^2+x+\dfrac1x-\dfrac1{x^2} \).
2. \( \dfrac{dy}{dx}=2x+1-\dfrac1{x^2}+\dfrac2{x^3} \).

Teacher's Note:
1. Expanding is quicker than the product rule here.
2. The derivative of \( -x^{-2} \) is \( +2x^{-3} \).

Answer: (a) 12 m/s

Explanation:
1. \( v=\dfrac{dx}{dt}=3t^2-6t+12 \).
2. At t = 0, \( v=12 \) m/s.

Teacher's Note:
1. Initial means t = 0.
2. 20 m is the initial position.

Answer: (d) −6 m/s²

Explanation:
1. \( a=\dfrac{dv}{dt}=6t-6 \).
2. At t = 0, \( a=-6 \) m/s².

Teacher's Note:
1. Differentiate the velocity again.
2. The particle is initially decelerating.

Answer: (c) 9 m/s

Explanation:
1. \( a=6t-6=0 \) gives t = 1 s.
2. \( v=3-6+12=9 \) m/s.

Teacher's Note:
1. Find the time first.
2. The velocity is a minimum at this instant, since a changes sign from negative to positive.

Answer: (b) \( 2\hat i+5\hat j+4\hat k \)

Explanation:
1. Add the components: \( (2+0)\hat i+(2+3)\hat j+(0+4)\hat k \).

Teacher's Note:
1. Add component by component.
2. Magnitude of the resultant is \( \sqrt{45} \) N.

Answer: (a) \( \cos^{-1}\dfrac{3}{5\sqrt2} \)

Explanation:
1. \( \vec F_1\cdot\vec F_2=0+6+0=6 \), \( |\vec F_1|=2\sqrt2 \), \( |\vec F_2|=5 \).
2. \( \cos\theta=\dfrac{6}{10\sqrt2}=\dfrac3{5\sqrt2} \).

Teacher's Note:
1. Use \( \cos\theta=\dfrac{\vec A\cdot\vec B}{AB} \).
2. \( |\vec F_1|=\sqrt{4+4} \).

Answer: (d) 6/5 N

Explanation:
1. Component \( =\vec F_1\cdot\hat F_2=\dfrac{6}{5} \) N.

Teacher's Note:
1. Dot with the unit vector of \( \vec F_2 \).
2. \( \hat F_2=(3\hat j+4\hat k)/5 \).

Answer: (c) Statement-1 is true and Statement-2 is false

Explanation:
1. Statement-1 is the definition of a vector.
2. Statement-2 is false: for example 5 N and 3 N in opposite directions give a resultant of 2 N, less than either.

Teacher's Note:
1. Test a statement with opposite vectors.
2. The resultant can be anywhere from |a − b| to a + b.

Answer: (b) Both statements are true but Statement-2 is not a correct explanation of Statement-1

Explanation:
1. Statement-1: \( \sqrt{8^2+6^2}=10 \) N, true.
2. Statement-2: \( \sin^2\theta+\cos^2\theta=1 \), true.
3. The two statements are unrelated, so Statement-2 does not explain Statement-1.

Teacher's Note:
1. Check each statement independently.
2. Explanation needs a logical link.

Answer: (a) Both statements are true and Statement-2 is a correct explanation of Statement-1

Explanation:
1. Statement-2 shows A is normal to the plane of B and C.
2. \( \vec B\times\vec C \) is also normal to that plane, so A is parallel to it: Statement-1 follows.

Teacher's Note:
1. Both vectors are normal to the same plane.
2. Statement-2 gives the geometric reason.

Answer: (d) Both statements are true and Statement-2 is a correct explanation of Statement-1

Explanation:
1. Two vectors of unequal magnitude cannot cancel, so at least three are needed.
2. Three unequal vectors forming the sides of a triangle in order do sum to zero, which explains Statement-1.

Teacher's Note:
1. A scalene triangle supplies the example.
2. Two vectors cancel only if equal and opposite.

Answer: (c) Both statements are true and Statement-2 is a correct explanation of Statement-1

Explanation:
1. \( (\hat i+\hat j)\cdot\hat k=0 \), so the angle is \( \cos^{-1}0=\pi/2 \).
2. Statement-2 is the formula used to get it.

Teacher's Note:
1. A zero dot product means perpendicular vectors.
2. Statement-2 is the general definition.

Answer: (b) Both statements are true and Statement-2 is a correct explanation of Statement-1

Explanation:
1. Length has magnitude only, so distance is a scalar.
2. Statement-2 says why: it is just a path length with no direction.

Teacher's Note:
1. A scalar has magnitude only.
2. Displacement is the vector counterpart.

Answer: (a) (1) and (3) only

Explanation:
1. The dot product is commutative: (1) true.
2. \( \vec B\times\vec A=-\vec A\times\vec B \), which differs unless the product is zero: (2) false.
3. \( AB\sin\theta=0 \) with non-zero magnitudes means \( \theta=0 \) or \( \pi \): (3) true.

Teacher's Note:
1. The cross product is anti-commutative.
2. Zero cross product means parallel or anti-parallel.

Answer: (d) \( \sqrt3\,a \)

Explanation:
1. The vectors are \( a\hat i+a\hat j+a\hat k \).
2. Magnitude \( =a\sqrt{1+1+1}=\sqrt3\,a \), the length of the body diagonal.

Teacher's Note:
1. The edges are mutually perpendicular.
2. The sum is the body diagonal.

Answer: (c) \( 15\hat i+20\hat j \)

Explanation:
1. \( |\vec B|=\sqrt{49+576}=25 \) and \( \hat A=\dfrac{3\hat i+4\hat j}5 \).
2. Required vector \( =25\hat A=15\hat i+20\hat j \).

Teacher's Note:
1. Scale the unit vector of A by 25.
2. The length 15² + 20² = 625.

Answer: (b) \( \sqrt{224} \) square units

Explanation:
1. \( \vec A\times\vec B=\begin{vmatrix}\hat i&\hat j&\hat k\\3&2&0\\2&0&-4\end{vmatrix}=-8\hat i+12\hat j-4\hat k \).
2. Area \( =\sqrt{64+144+16}=\sqrt{224}\approx14.97 \).

Teacher's Note:
1. The area is the magnitude of the cross product.
2. Note that \( \sqrt{224}\ne14 \).

Answer: (a) \( 45^\circ \)

Explanation:
1. \( \tan\theta=\dfrac11 \), so \( \theta=45^\circ \).

Teacher's Note:
1. Equal components give 45°.
2. \( \cos\theta=1/\sqrt2 \).

Answer: (d) 2

Explanation:
1. \( |\vec A+\vec B|^2=2\alpha^2(1+\cos\omega t) \) and \( |\vec A-\vec B|^2=2\alpha^2(1-\cos\omega t) \).
2. \( 1+\cos\omega t=3(1-\cos\omega t) \), so \( \cos\omega t=\tfrac12 \).
3. First time: \( \omega\tau=\dfrac\pi3 \), so \( \tau=\dfrac{\pi/3}{\pi/6}=2 \) s.

Teacher's Note:
1. Square both sides and use the dot product.
2. The first solution of cos = 1/2 is π/3.

Answer: (c) I-q, II-r, III-p, IV-s

Explanation:
1. I: \( A^2=2A^2(1+\cos\theta) \) gives \( \cos\theta=-\tfrac12 \), so \( 120^\circ \).
2. II: the resultant lies between 4 N and 12 N, so 12 N is possible.
3. III: the dot product is 0, so \( 90^\circ \). IV: \( |2\hat i+\hat j+3\hat k|=\sqrt{14} \).

Teacher's Note:
1. Check each pairing separately.
2. The resultant can equal 12 N only for parallel forces.