JEE Physics Basic Mathematics Used in Physics MCQs Set 02

Answer: (b) 3

Explanation:
1. Space has three mutually perpendicular axes (x, y and z).
2. So at most 3 rectangular components.

Teacher's Note:
1. Three perpendicular axes exist in space.
2. Use î, ĵ and k̂.

Answer: (b) 8 N

Explanation:
1. The resultant lies between 10 − 6 = 4 N and 10 + 6 = 16 N.
2. Only 8 N lies in this range.

Teacher's Note:
1. The magnitude range is
2. A − B
3. to A + B.
4. Check each option against it.

Answer: (d) 2 N

Explanation:
1. The resultant lies between 4 N and 16 N.
2. 2 N is outside this range, so it cannot be obtained.

Teacher's Note:
1. Use the minimum and maximum possible resultants.
2. The minimum is the difference of the magnitudes.

Answer: (d) 1 N and 4 N

Explanation:
1. For 1 N and 4 N the resultant lies between 3 N and 5 N, so 2 N is impossible.
2. The other pairs allow 2 N (2, 2: 0 to 4; 1, 1: 0 to 2; 1, 3: 2 to 4).

Teacher's Note:
1. Check whether 2 N is within the range for each pair.
2. The range is the difference to the sum.

Answer: (a) 0

Explanation:
1. The magnitude of the sum equals the sum of the magnitudes only when the vectors are parallel in the same direction.
2. So the angle is 0.

Teacher's Note:
1. C = A + B is the maximum resultant.
2. It occurs at 0°.

Answer: (c) 2(A² + B²)

Explanation:
1. R₁² = A² + B² + 2AB cos θ and R₂² = A² + B² − 2AB cos θ.
2. Adding gives R₁² + R₂² = 2(A² + B²).

Teacher's Note:
1. Reversing a vector changes cos θ to −cos θ.
2. The cross terms cancel.

Answer: (b) \( \vec Q=\vec 0 \)

Explanation:
1. Cancelling \( \vec P \) from both sides gives \( \vec Q=-\vec Q \), so \( \vec Q=\vec0 \).

Teacher's Note:
1. Simplify the equation algebraically.
2. A vector equal to its negative is the zero vector.

Answer: (c) F₁ = 3 N, F₂ = 5 N, F₃ = 6 N

Explanation:
1. For equilibrium the third force must lie between
2. 5 − 3
3. = 2 N and 5 + 3 = 8 N.
4. Only 6 N does.

Teacher's Note:
1. Equilibrium means the three forces can form a closed triangle.
2. The third side must lie between the difference and the sum of the other two.

Answer: (d) \( \sqrt3|\vec A| \)

Explanation:
1. @
2. \vec A+\vec B
3. =A\( with A = B gives \)A²(2+2\cos\theta)=A²\( , so \)\cos\theta=-\tfrac12@ and θ = 120°.
4. @
5. \vec A-\vec B
6. ²=2A²(1-\cos\theta)=3A²\( , so it equals \)\sqrt3\,A@.

Teacher's Note:
1. First find the angle between the vectors.
2. Then use the formula for the difference.

Answer: (b) direction reverses and magnitude is doubled

Explanation:
1. A negative scalar reverses the direction.
2. The magnitude is multiplied by 2, and the unit stays the same.

Teacher's Note:
1. Multiplying by a scalar changes the magnitude by its modulus.
2. The sign decides the direction.

Answer: (d) 120°

Explanation:
1. R² = A² + A² + 2A² cos θ = A², so \( \cos\theta=-\tfrac12 \).
2. θ = 120°.

Teacher's Note:
1. Equal magnitudes: R = 2A cos(θ/2).
2. Setting R = A gives cos(θ/2) = 1/2.

Answer: (b) \( \sqrt3 \)

Explanation:
1.
2. A + B
3. ² = 2 + 2cos θ = 1, so \( \cos\theta=-\tfrac12 \) (θ = 120°).
4.
5. A − B
6. ² = 2 − 2cos θ = 3, so the difference has magnitude \( \sqrt3 \).

Teacher's Note:
1. Find the angle first.
2. The difference uses 1 − cos θ.

Answer: (b) \( \sqrt{0.11} \)

Explanation:
1. A unit vector has magnitude 1: 0.25 + 0.64 + c² = 1.
2. c² = 0.11, so c = √0.11.

Teacher's Note:
1. The sum of the squares of the components is 1.
2. Subtract the known squares from 1.

Answer: (c) 2

Explanation:
1. cos²α + cos²β + cos²γ = 1 for direction cosines.
2. So sin²α + sin²β + sin²γ = 3 − 1 = 2.

Teacher's Note:
1. The direction cosines satisfy l² + m² + n² = 1.
2. Use sin² = 1 − cos².

Answer: (c) \( \frac12,\frac12,\frac1{\sqrt2} \)

Explanation:
1. The magnitude is \( \sqrt{1+1+2}=2 \).
2. The direction cosines are the components divided by 2: \( \frac12,\frac12,\frac{\sqrt2}{2}=\frac1{\sqrt2} \).

Teacher's Note:
1. Divide each component by the magnitude.
2. The squares of the cosines add to 1.

Answer: (c) \( \frac{\hat i+\hat j}{\sqrt2} \)

Explanation:
1. The magnitude of \( \hat i+\hat j \) is \( \sqrt2 \).
2. The unit vector is \( \frac{\hat i+\hat j}{\sqrt2} \).

Teacher's Note:
1. Divide by the magnitude.
2. A unit vector has magnitude 1.

Answer: (a) \( \frac17(3\hat i+6\hat j-2\hat k) \)

Explanation:
1. The resultant is \( 3\hat i+6\hat j-2\hat k \), with magnitude \( \sqrt{9+36+4}=7 \).
2. The unit vector is \( \frac17(3\hat i+6\hat j-2\hat k) \).

Teacher's Note:
1. Add the components first.
2. Divide by the magnitude, not its square.

Answer: (b) \( \hat j-\hat k \)

Explanation:
1. \( \vec B=\hat i-\vec A=\hat i-(\hat i-\hat j+\hat k)=\hat j-\hat k \).

Teacher's Note:
1. Subtract A from the resultant.
2. The unit vector along x is î.

Answer: (b) 13 N

Explanation:
1. For mutually perpendicular forces: R = √(3² + 4² + 12²) = √169 = 13 N.

Teacher's Note:
1. Use the three-dimensional Pythagoras theorem.
2. 3, 4, 12, 13 is a Pythagorean quadruple-like set.

Answer: (b) \( \tan^{-1}(2/3) \)

Explanation:
1. The y-component is 3 and the x-component is 2.
2. With the y-axis, \( \tan\phi=\frac{A_x}{A_y}=\frac23 \), so \( \phi=\tan^{-1}(2/3) \).

Teacher's Note:
1. The angle with the y-axis uses the x-component in the numerator.
2. Draw the vector to be sure.

Answer: (a) 0°

Explanation:
1. The resultant is \( (\vec A+\vec B)+(\vec A-\vec B)=2\vec A \).
2. This is parallel to \( \vec A \), so the angle is 0°.

Teacher's Note:
1. Add the two vectors algebraically first.
2. Vector 2A points along A.

Answer: (d) \( \frac1{\sqrt2},-\frac1{\sqrt2} \)

Explanation:
1. Perpendicular: a + b = 0, so b = −a.
2. For a unit vector, a² + b² = 1, so \( a=\frac1{\sqrt2},\ b=-\frac1{\sqrt2} \).

Teacher's Note:
1. The dot product must be zero.
2. Only option 4 satisfies a + b = 0.

Answer: (c) 90°

Explanation:
1. \( (\vec A+\vec B)\cdot(\vec A-\vec B)=A^2-B^2=0 \) when A = B.
2. So they are perpendicular.

Teacher's Note:
1. The diagonals of a rhombus are perpendicular.
2. Use the dot product.

Answer: (b) are equal to each other in magnitude

Explanation:
1. \( (\vec P+\vec Q)\cdot(\vec P-\vec Q)=P^2-Q^2=0 \), so P = Q.
2. They are equal in magnitude, not necessarily equal as vectors.

Teacher's Note:
1. Use the dot product of the sum and the difference.
2. Equal magnitude does not mean equal vectors.

Answer: (c) \( \pi/2 \)

Explanation:
1. 12² + 5² = 144 + 25 = 169 = 13², so C² = A² + B².
2. Hence cos θ = 0 and θ = π/2.

Teacher's Note:
1. The numbers form a Pythagorean triple.
2. Perpendicular vectors satisfy C² = A² + B².

Answer: (c) \( \cos^{-1}\frac{12}{13} \)

Explanation:
1. Since 5² + 12² = 13², P and Q are perpendicular and R is the hypotenuse.
2. The angle between Q and R has cos = Q/R = 12/13.

Teacher's Note:
1. Draw the right triangle of vectors.
2. The adjacent side to the angle between Q and R is Q.

Answer: (b) \( 7\hat k \)

Explanation:
1. The dot product with \( 7\hat k \) is zero.
2. The other options give 7, 24 and 24 respectively.

Teacher's Note:
1. Perpendicular vectors have zero dot product.
2. A vector along z is perpendicular to any vector in the xy-plane.

Answer: (a) zero

Explanation:
1. W = (3)(2) + (2)(−3) = 6 − 6 = 0.

Teacher's Note:
1. Work is the dot product of force and displacement.
2. The force and displacement are perpendicular.

Answer: (a) 0°

Explanation:
1. PQ cos θ = PQ gives cos θ = 1, so θ = 0°.

Teacher's Note:
1. The dot product is largest for parallel vectors.
2. Compare with the definition.