This problem involves understanding relative directions. We need to visualize the positions of points 1, 2, and 3 based on the given information and then determine the direction of point 2 with respect to point 3.
C) North east. As explained in the step-by-step analysis, from point 3, point 2 is located in the North-East direction.
The question asks about the fundamental concept behind spherical lenses. We need to identify the primary optical phenomenon that lenses utilize to form images.
B) Light refraction — Spherical lenses work on the principle of light refraction. When light passes through a lens, it changes speed and direction as it enters and exits the lens material, causing it to bend. This bending of light is called refraction, and it is what allows lenses to converge or diverge light rays to form images.
The question asks to identify the Indian city known as the 'City of Joy'. This is a general knowledge question related to popular nicknames of Indian cities. We need to recall the common epithets associated with the given options.
Correct Option: D) Kolkata is famously known as the 'City of Joy'. This nickname was popularized by Dominique Lapierre's 1985 novel of the same name, which depicted life in a Calcutta (now Kolkata) slum. The city is celebrated for its rich cultural heritage, intellectual vibrancy, and festive spirit.
Rs.16,500 is split into two fixed deposits for 2 years at 10% and 12% Simple Interest. If the total interest received from the deposits is Rs.3,620, how much was invested in the 12% scheme?
This problem involves simple interest calculations for two different investments. We are given the total principal amount, the interest rates for two schemes, the time period, and the total interest received. We need to find the amount invested in one of the schemes. We can set up equations based on the simple interest formula and solve them simultaneously.
Let the amount invested in the 10% scheme be \(P_1\) and the amount invested in the 12% scheme be \(P_2\).
The total principal amount is Rs. 16,500. So, we have:
\[ P_1 + P_2 = 16500 \quad \text{(Equation 1)} \]The formula for Simple Interest (SI) is:
\[ \text{SI} = \frac{\text{Principal} \times \text{Rate} \times \text{Time}}{100} \]For the first scheme (10% interest rate, 2 years time):
\[ \text{SI}_1 = \frac{P_1 \times 10 \times 2}{100} = \frac{20 P_1}{100} = \frac{P_1}{5} \]For the second scheme (12% interest rate, 2 years time):
\[ \text{SI}_2 = \frac{P_2 \times 12 \times 2}{100} = \frac{24 P_2}{100} = \frac{6 P_2}{25} \]The total interest received is Rs. 3,620. So, we have:
\[ \text{SI}_1 + \text{SI}_2 = 3620 \] \[ \frac{P_1}{5} + \frac{6 P_2}{25} = 3620 \quad \text{(Equation 2)} \]Now we have a system of two linear equations:
Equation 1: \(P_1 + P_2 = 16500\)
Equation 2: \[\frac{P_1}{5} + \frac{6 P_2}{25} = 3620\]
From Equation 1, express \(P_1\) in terms of \(P_2\):
\[ P_1 = 16500 - P_2 \]Substitute this expression for \(P_1\) into Equation 2:
\[ \frac{(16500 - P_2)}{5} + \frac{6 P_2}{25} = 3620 \]Multiply the entire equation by 25 to eliminate denominators:
\[ 5(16500 - P_2) + 6 P_2 = 3620 \times 25 \] \[ 82500 - 5 P_2 + 6 P_2 = 90500 \]Simplify and solve for \(P_2\):
\[ 82500 + P_2 = 90500 \] \[ P_2 = 90500 - 82500 \] \[ P_2 = 8000 \]The amount invested in the 12% scheme (\(P_2\)) is Rs. 8,000.
C) Rs. 8,000 is the amount invested in the 12% scheme, as calculated above.
The question asks to identify the field with which Shubha Mudgal is associated. This requires general knowledge about prominent Indian personalities and their respective professions or fields of expertise.
Correct Option: C) Shubha Mudgal is a renowned Indian classical vocalist who specializes in the Hindustani classical tradition. She is celebrated for her powerful voice and versatile repertoire, including Khayal, Thumri, and Dadra.
The speed of light in a medium is inversely proportional to its refractive index. This means that a lower refractive index corresponds to a higher speed of light, and vice versa. We need to identify the substance with the lowest refractive index to find where the speed of light is maximum.
Correct Option: D) A
To find out when the bells will toll together again, we need to find the least common multiple (LCM) of their individual tolling intervals. Once we have the LCM, which represents the time after which they will toll together, we can determine how many times they will toll together within the given total duration of 30 minutes.
Correct Option: B) 16
This question tests general knowledge about famous personalities, specifically nicknames of well-known cricketers. To answer correctly, one needs to recall the common nickname associated with Ricky Ponting.
Correct Option: C) Punter
This problem involves calculating the shortest distance from the starting point after a series of movements in different directions. This can be solved by visualizing the movements and using the Pythagorean theorem, as the final displacement forms the hypotenuse of a right-angled triangle.
C) 10 km is the correct distance from the starting point.
To determine the number of electrons in one Coulomb of charge, we need to use the fundamental charge of a single electron. The total charge is given as one Coulomb, and the charge of one electron is a known physical constant. We can find the number of electrons by dividing the total charge by the charge of a single electron.
Identify the given total charge: The total charge \(Q\) is given as one Coulomb.
\[ Q = 1 \; \text{C} \]Recall the charge of a single electron: The charge of one electron, denoted as \(e\), is a fundamental constant.
\[ e = 1.602 \times 10^{-19} \; \text{C} \]Calculate the number of electrons: The number of electrons (\(n\)) can be found by dividing the total charge by the charge of a single electron.
\[ n = \frac{Q}{e} = \frac{1 \; \text{C}}{1.602 \times 10^{-19} \; \text{C}} \]Perform the division:
\[ n = \frac{1}{1.602 \times 10^{-19}} = \frac{1}{1.602} \times 10^{19} \] \[ n \approx 0.6242 \times 10^{19} \]Express in standard scientific notation: To express this in standard scientific notation (with one non-zero digit before the decimal point), we adjust the decimal place.
\[ n \approx 6.242 \times 10^{18} \]Rounding to two decimal places, we get:
\[ n \approx 6.25 \times 10^{18} \; \text{electrons} \]Correct Option: C) 6.25 × 10¹⁸ electrons