Your Answer : (Not Answered)

Correct Answer : D

The combination that does not work is 6, 3.

\[ 6 \times 200 + 3 \times 300 = 1200 + 900 = 2100\ \text{mL} \]

That overflows the 2000 mL bucket. The other combinations fill it exactly: 1 × 200 + 6 × 300 = 2000; 4 × 200 + 4 × 300 = 2000; and 7 × 200 + 2 × 300 = 2000.

Your Answer : (Not Answered)

Correct Answer : A

The factorisation is (3x - 2y + 3)(3x - 2y + 4).

Group the terms around \(t = 3x - 2y\):

\[ \begin{aligned} 9x^{2} - 12xy + 4y^{2} &= (3x - 2y)^{2} = t^{2} \\ 21x - 14y &= 7(3x - 2y) = 7t \end{aligned} \]

So the expression is a quadratic in \(t\):

\[ \begin{aligned} t^{2} + 7t + 12 &= (t + 3)(t + 4) \\ &= (3x - 2y + 3)(3x - 2y + 4) \end{aligned} \]

Options with \(3x + 2y\) would give \(+12xy\) and \(+14y\), the wrong signs. (t + 2)(t + 6) multiplies out to \(t^{2} + 8t + 12\), not \(t^{2} + 7t + 12\).

Your Answer : (Not Answered)

Correct Answer : C

The ratio is 21 : 22. Let the numbers be \(10k\) and \(13k\). \[ \frac{10k-90}{13k-90} = \frac{7}{10} \] \[ 100k - 900 = 91k - 630 \] \[ 9k = 270,\quad k = 30 \] So A = 300 and B = 390. Adding 57 to A and subtracting 16 from B: \[ A = 357,\quad B = 374 \] \[ 357 : 374 = 21 : 22 \] since both are divisible by 17. The ratios 19 : 20, 20 : 21 and 22 : 23 do not match 357 : 374. Check: with \(k = 30\) the numbers are 300 and 390, and after subtracting 90 they become 210 and 300, which is 7 : 10 as the question says.

Your Answer : (Not Answered)

Correct Answer : B

We can multiply sequentially.

\(0.5 \times 0.05 = 0.025\).

Now, \(0.025 \times 500 = 25 \times 0.5 = 12.5\).

Alternatively, write as fractions: \((\frac{1}{2}) \times (\frac{5}{100}) \times 500 = (\frac{1}{2}) \times (\frac{1}{20}) \times 500 = \frac{500}{40} = \frac{50}{4} = 12.5\).

Your Answer : (Not Answered)

Correct Answer : A

This expression is in the form of the cosine addition formula: \(\cos(A+B) = \cos A \cos B - \sin A \sin B\).

Here, A = 38° and B = 52°.

The expression is equal to \(\cos(38° + 52°) = \cos(90°)\).

The value of \(\cos 90°\) is 0.

Your Answer : (Not Answered)

Correct Answer : B

If the right angle is at B, then the sides AB and BC are perpendicular. The product of their slopes must be -1.

Slope of AB = \(\frac{3-7}{p-4} = \frac{-4}{p-4}\).

Slope of BC = \(\frac{3-3}{7-p} = \frac{0}{7-p} = 0\).

A slope of 0 means BC is a horizontal line. For AB to be perpendicular to a horizontal line, it must be a vertical line. A vertical line has an undefined slope, which happens when its denominator is 0.

So, p - 4 = 0, which means p = 4.

Your Answer : (Not Answered)

Correct Answer : A

Both aircraft are on the same vertical line, so they share one horizontal distance \( d \) from the observer.

For the upper plane at 900 m, elevation 60°:

\[ \tan 60^{\circ} = \frac{900}{d} \quad\Longrightarrow\quad d = \frac{900}{\sqrt{3}} = 300\sqrt{3} = 519.6 \text{ m} \]

For the lower plane, elevation 45°:

\[ h = d\tan 45^{\circ} = 519.6 \text{ m} \]

\[ \text{difference} = 900 - 519.6 = 380.4 \approx 381 \text{ m} \]

The steeper 60° sighting must belong to the higher aircraft, which fixes which angle goes with which plane.

Your Answer : (Not Answered)

Correct Answer : B

The total possible outcomes = 24

Numbers divisible by only 2 are 2, 4, 8, 10, 14, 16, 20, 22 (8 numbers) --- (1)

Numbers divisible only by 3 are 3, 9, 15, 21 (4 numbers) --- (2)

Numbers divisible by both 2 and 3 are 6, 12, 18, 24 (4 numbers) --- (3)

From (1), (2) and (3), we see that the number of favourable outcomes is 16 (i.e., 8 + 4 + 4).

We know that, Probability of an event E, P (E) =number of favourable outcomes/ total number of outcomes

=16/24

=2/3

Your Answer : (Not Answered)

Correct Answer : C

A sector is a triangle-like wedge whose "base" is its arc and whose "height" is the radius:

\[ A = \frac{1}{2} \times r \times \text{arc length} \]

\[ = \frac{1}{2} \times 5 \times 3.5 = 8.75 \text{ cm}^{2} \]

This form needs no angle at all. It follows from \( A = \tfrac{\theta}{2\pi}\pi r^{2} \) and \( \text{arc} = r\theta \), which combine to give exactly \( \tfrac{1}{2}r \times \text{arc} \).

Your Answer : (Not Answered)

Correct Answer : B

Fencing follows the circumference, and the diameter is given directly.

\[ C = \pi d = \frac{22}{7} \times 28 = 88 \text{ m} \]

\[ \text{cost} = 88 \times 1.50 = \text{Rs. } 132 \]

Using \( 2\pi d \) - effectively treating 28 as the radius - would double the answer to Rs. 264.

Your Answer : (Not Answered)

Correct Answer : D

A trapezium's area is the average of the parallel sides times the depth between them.

\[ A = \frac{1}{2}(a + b)h \]

\[ 640 = \frac{1}{2}(10 + 6)h = 8h \]

\[ h = 80 \text{ m} \]

The two widths must be averaged, not added outright: using 16 as the effective width would halve the depth to 40 m.

Your Answer : (Not Answered)

Correct Answer : A

P= Rs 900

Rate = 4.5*

SI= Rs 81

=> SI= Time*P*rate

calculating for rate,

=> Time= SI/rate*P

= 81/4.5%*900

=> 2 years

Thus, it will take 2 years to yield Rs 81.

Your Answer : (Not Answered)

Correct Answer : A

Correct answer: 12 seconds

Moving in opposite directions, the speeds add.

\[54 + 36 = 90 \text{ km/hr} = 90 \times \frac{5}{18} = 25 \text{ m/s}\]

They must together cover the sum of their lengths, 300 m.

\[t = \frac{300}{25} = 12 \text{ seconds}\]

Your Answer : (Not Answered)

Correct Answer : D

Since the month begins with a Sunday, to there will be five Sundays in the month.
Required average
\(\left( \dfrac {510 \times 5 + 240 \times 25}{30} \right)\\ =\dfrac{8550}{30}\\ =285\)

Your Answer : (Not Answered)

Correct Answer : B

The collector current is 1.5 mA.

The current gain \(\beta\) relates collector current to base current:

\[ \begin{aligned} I_{C} &= \beta I_{B} = 50 \times 30\ \mu\text{A} \\ &= 1500\ \mu\text{A} = 1.5\ \text{mA} \end{aligned} \]

15 mA and 150 mA come from slipping the decimal point by one or two places when converting microamperes to milliamperes, and 1.5 A treats the base current as 30 mA. A useful check is that small-signal transistors biased with tens of microamperes of base current carry collector currents of a few milliamperes.

Your Answer : (Not Answered)

Correct Answer : C

The correct matching is a-ii, b-iv, c-iii, d-i.

• Ampere's law, with Maxwell's displacement current: \(\nabla \times \bar{H} = \bar{J} + \partial\bar{D}/\partial t\) (ii).
• Faraday's law: \(\nabla \times \bar{E} = -\partial\bar{B}/\partial t\) (iv).
• Gauss' law for electric fields: \(\nabla \cdot \bar{D} = \rho\) (iii).
• No magnetic monopoles: \(\nabla \cdot \bar{B} = 0\) (i).

The curl equations describe circulation (Ampere and Faraday) and the divergence equations describe sources (charge, and the absence of magnetic charge). The near-miss code swaps the two divergence equations, giving the monopole law the charge density.

Your Answer : (Not Answered)

Correct Answer : B

The law is exact for any closed path whatever. Symmetry is needed only to extract the field from under the integral sign: without it the line integral still equals the enclosed current but the field cannot be solved for.

Your Answer : (Not Answered)

Correct Answer : B

The integral is zero.

The integrand \(1/(z+2)\) has a single simple pole, at \(z = -2\). The contour is the unit circle \(|z| = 1\), and since \(|-2| = 2 \gt 1\) the pole lies outside it. Inside and on the contour the function is analytic, so by Cauchy's theorem

\[ \oint_{|z|=1} \frac{dz}{z+2} = 0 \]

The value \(2\pi i\) would be the answer only if the pole were enclosed (residue 1 times \(2\pi i\)); \(-2\pi i\) is that with the contour traversed clockwise, and \(\pi i\) arises for a pole lying on a contour, in the principal-value sense. None of these applies here.

Your Answer : (Not Answered)

Correct Answer : A

The tensor product is \(\frac{1}{2}(1, -i, i, 1)^{T}\), a column vector.

For column vectors \(a\) and \(b\), the tensor product is the column \((a_{1}b_{1}, a_{1}b_{2}, a_{2}b_{1}, a_{2}b_{2})^{T}\). With \(a = (1, i)\), \(b = (1, -i)\) and the prefactor \(\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = \frac{1}{2}\):

\[ \begin{aligned} a_{1}b_{1} &= 1, \quad a_{1}b_{2} = -i \\ a_{2}b_{1} &= i, \quad a_{2}b_{2} = (i)(-i) = 1 \end{aligned} \]

The option \((1, i, -i, 1)\) is \(|v_{2}\rangle \otimes |v_{1}\rangle\), the product taken in the other order. The two 2 by 2 matrices are outer products of the form \(|v\rangle\langle w|\), not the tensor product of two kets, which is a four-component column.

Your Answer : (Not Answered)

Correct Answer : B

The real engine satisfies \(\eta \lt \eta_{C}\).

The Carnot efficiency \(\eta_{C} = 1 - T_{min}/T_{max}\) is the highest possible for any engine working between these temperature limits. A real engine falls short for two reasons: it is irreversible, and its heat is added and rejected over a range of temperatures, so the mean temperature of heat addition is below \(T_{max}\) and the mean temperature of rejection is above \(T_{min}\).

\(\eta \gt \eta_{C}\) would violate the second law, \(\eta = \eta_{C}\) needs a fully reversible engine with heat exchange only at the extremes, and \(1 + \eta_{C}\) exceeds 100%.