The question asks to identify the technology used by NCRA-TIFR astronomers in February 2022 for mapping atomic hydrogen gas distribution in a galaxy. This requires knowledge of recent astronomical research and the capabilities of various telescopes/satellites.
Correct Option: C) The Giant Metrewave Radio Telescope (GMRT) is an array of thirty parabolic radio telescopes located near Pune, India. It is operated by the National Centre for Radio Astrophysics (NCRA) of the Tata Institute of Fundamental Research (TIFR). GMRT is specifically designed to observe radio waves in the metre wavelength range, which includes the 21 cm line emission from neutral atomic hydrogen (\(\text{HI}\)) gas. In February 2022, astronomers from NCRA-TIFR indeed used the upgraded GMRT (uGMRT) to map the distribution of atomic hydrogen gas in a galaxy called "The Tadpole Galaxy" (UGC 10214), providing unprecedented details about its tidal tails and star formation regions. This aligns perfectly with the question's premise.
The question asks about India's trade policy during its first seven Five-Year Plans. This requires knowledge of India's economic history and the strategies adopted for industrialization and self-reliance post-independence.
Correct Option: D) Import substitution. India adopted a strategy of Import Substitution Industrialization (ISI) during its first seven Five-Year Plans to promote domestic industries and achieve self-reliance. This involved protecting domestic industries from foreign competition through various trade barriers.
To find the total distance covered in 4 hours, we first need to determine the total distance of the journey described in the problem. We can set up an equation for the total time taken for the journey based on the given speeds and fractions of distance. Once the total distance is found, we can calculate the distance covered in 4 hours.
Let the total distance be \(D\) km.
One-third of the distance is \(\frac{D}{3}\) km. This part was covered at a speed of 30 km/h.
Time taken for the first part (\(t_1\)) is:
\[ t_1 = \frac{\text{Distance}}{\text{Speed}} = \frac{D/3}{30} = \frac{D}{90} \text{ hours} \]The remaining distance is \[D - \frac{D}{3} = \frac{2D}{3}\] km. This part was covered at a speed of 20 km/h.
Time taken for the second part (\(t_2\)) is:
\[ t_2 = \frac{\text{Distance}}{\text{Speed}} = \frac{2D/3}{20} = \frac{2D}{60} = \frac{D}{30} \text{ hours} \]The total time taken for the entire journey (let's call it \(T_{\text{total}}\)) is the sum of \(t_1\) and \(t_2\).
\[ T_{\text{total}} = t_1 + t_2 = \frac{D}{90} + \frac{D}{30} \]To add these fractions, find a common denominator, which is 90:
\[ T_{\text{total}} = \frac{D}{90} + \frac{3D}{90} = \frac{D + 3D}{90} = \frac{4D}{90} = \frac{2D}{45} \text{ hours} \]The problem states that the car covered a certain distance in 4 hours. This implies that the total time \(T_{\text{total}}\) is 4 hours for the distance we are trying to find.
So, we set \(T_{\text{total}} = 4\):
\[ \frac{2D}{45} = 4 \]Now, solve for \(D\):
\[ 2D = 4 \times 45 \] \[ 2D = 180 \] \[ D = \frac{180}{2} \] \[ D = 90 \text{ km} \]This means that the total distance covered by the car in 4 hours is 90 km.
C) 90 km is the correct answer. The calculation shows that under the given conditions, the car covers 90 km in 4 hours.
A) 60 km — This might be obtained if there was an error in calculating the total time or if the speeds were averaged incorrectly without considering the different distances covered at each speed.
B) 30 km — This value is too low and does not align with the given speeds and time. It might result from considering only one segment of the journey or a miscalculation.
D) 95 km — This value is close to the correct answer but is not precise. It could be a result of a minor calculation error during the process of finding the total distance.
The question asks to identify the Indian women's cricketer who was awarded the ICC Women's Cricketer of the Year for her performance in 2021. This requires knowledge of major cricket awards and their recipients.
Correct Option: C) Smriti Mandhana was named the ICC Women's Cricketer of the Year (Rachael Heyhoe Flint Trophy) for 2021. She had an exceptional year across all formats, scoring 855 runs in 22 international matches at an average of 38.86, including one century and five half-centuries.
To determine the value of \(k\) for which a pair of linear equations has no solution, we need to recall the conditions for consistency of linear equations. For a pair of linear equations given by \(a_1x + b_1y + c_1 = 0\) and \(a_2x + b_2y + c_2 = 0\), there is no solution if the lines are parallel and distinct. This condition is met when the ratio of the coefficients of \(x\) and \(y\) are equal, but not equal to the ratio of the constant terms.
Identify the given linear equations:
\[ 2x + 3y - 1 = 0 \quad \text{(Equation 1)} \]\[ 4x + ky + 2 = 0 \quad \text{(Equation 2)} \]Compare these equations with the general form \(a_1x + b_1y + c_1 = 0\) and \(a_2x + b_2y + c_2 = 0\).
From Equation 1:
\[ a_1 = 2, \quad b_1 = 3, \quad c_1 = -1 \]From Equation 2:
\[ a_2 = 4, \quad b_2 = k, \quad c_2 = 2 \]For a pair of linear equations to have no solution, the condition is:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \]Substitute the values into the first part of the condition:
\[ \frac{2}{4} = \frac{3}{k} \]Simplify the ratio on the left side:
\[ \frac{1}{2} = \frac{3}{k} \]Solve for \(k\):
\[ 1 \times k = 2 \times 3 \]\[ k = 6 \]Now, check the second part of the condition with \(k=6\):
\[ \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \]\[ \frac{3}{6} \neq \frac{-1}{2} \]\[ \frac{1}{2} \neq -\frac{1}{2} \]Since \[\frac{1}{2} \neq -\frac{1}{2}\], the condition for no solution is satisfied when \(k=6\).
Correct Option: A) 6
The problem asks us to find a set of numbers that follows the same relationship as the given sets: (21, 225, 9) and (4, 25, 6). We need to identify the pattern connecting the three numbers in each set.
Correct Option: C) (14, 64, 2)
The power of a lens is the reciprocal of its focal length. For a convex lens, the focal length is positive. The focal length is given in centimeters, so it must be converted to meters before calculating the power, as the unit of power (Dioptre, D) requires focal length in meters.
Correct Option: C) 2.5 D is the calculated power of the convex lens.
This question tests the properties of similar triangles, specifically how the ratio of their areas relates to the ratio of their corresponding linear elements (sides, altitudes, medians, angle bisectors). We need to identify the statement that is NOT true.
Let's consider two similar triangles, \(\triangle ABC \sim \triangle PQR\). If the ratio of their corresponding sides is \(k\), then:
Now, let's evaluate each option:
Since \[\frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle PQR)} = k^2\] and \(\frac{AE}{PT} = k\), it follows that \[\frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle PQR)} = \left(\frac{AE}{PT}\right)^2\]. This statement is TRUE.
This would mean \[\frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle PQR)} = \frac{AE}{PT}\]. But we know \[\frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle PQR)} = k^2\] and \(\frac{AE}{PT} = k\). Since \(k^2 \neq k\) (unless \(k=1\), which is a special case of congruent triangles, or \(k=0\), which is not possible), this statement is generally FALSE.
Since \[\frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle PQR)} = k^2\] and \(\frac{AM}{PN} = k\), it follows that \[\frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle PQR)} = \left(\frac{AM}{PN}\right)^2\]. This statement is TRUE.
Since \[\frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle PQR)} = k^2\] and \(\frac{AD}{PS} = k\), it follows that \[\frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle PQR)} = \left(\frac{AD}{PS}\right)^2\]. This statement is TRUE.
B) Ratio of areas of two similar triangles = Ratio of corresponding segments obtained by corresponding angle bisectors.
\(\frac{a^{10}\times b^{-7}\times c^{-4}}{a^{-5}\times b^{2}\times c^{5}}\) in simplified form is:
a¹⁵ × b⁹ × c⁹
a⁵ × b⁻⁹ × c⁻⁹
a⁵ × b⁹ × c⁻⁹
a¹⁵ × b⁻⁹ × c⁻⁹
To simplify the given expression, we will use the laws of exponents. Specifically, we will use the rule \(\frac{x^m}{x^n} = x^{m-n}\) for division of terms with the same base, and \(x^{-n} = \frac{1}{x^n}\) for negative exponents.
The given expression is:
\[ \frac{a^{10} \times b^{-7} \times c^{-4}}{a^{-5} \times b^{2} \times c^{5}} \]Group the terms with the same base:
\[ \left( \frac{a^{10}}{a^{-5}} \right) \times \left( \frac{b^{-7}}{b^{2}} \right) \times \left( \frac{c^{-4}}{c^{5}} \right) \]Apply the exponent rule \(\frac{x^m}{x^n} = x^{m-n}\) to each group:
Combine the simplified terms:
\[ a^{15} \times b^{-9} \times c^{-9} \]Correct Option: D) a¹⁵ × b⁻⁹ × c⁻⁹