An object of mass 10g requires 20J of heat to raise its temperature by 20°C. Calculate its specific heat capacity.

Here are the solutions to questions 12, 13, 14, 15, 16, 17, 18, 19, and 20. 12. An object of mass 10g requires 20J of heat to raise its temperature by 20°C. Calculate its specific heat capacity. Step 1: Identify the given values. Mass, m = 10 g Heat energy, Q = 20 J Temperature change, T = 20^ = 20 K Step 2: Use the formula for heat energy. Q = mc T where c is the specific heat capacity. Step 3: Rearrange the formula to solve for c. c = (Q)/(m T) Step 4: Substitute the values and calculate c. c = 20 J(10 g)(20 K) c = 20 J200 g K c = 0.1 J g^-1K^-1 The correct option is A. 0.1 J g^-1K^-1. 13. A tyre is pumped to a pressure of 30 Nm^-2 at 27^. When heated to 54^, find the new pressure (volume constant). Step 1: Convert temperatures from Celsius to Kelvin. Initial temperature, T_1 = 27^ + 273 = 300 K Final temperature, T_2 = 54^ + 273 = 327 K Step 2: Identify the initial pressure. Initial pressure, P_1 = 30 Nm^-2 Step 3: Use Gay-Lussac's Law for constant volume. (P_1)/(T_1) = (P_2)/(T_2) Step 4: Rearrange to solve for P_2. P_2 = P_1 (T_2)/(T_1) Step 5: Substitute the values and calculate P_2. P_2 = (30 Nm^-2) 327 K300 K P_2 = 30 × 1.09 Nm^-2 P_2 = 32.7 Nm^-2 The correct option is C. 32.7 Nm^-2. 14. How long will it take to heat 3kg of water from 28^ to 88^ using an electric kettle that draws 6A from a 210V supply? Assumption: Specific heat capacity of water c_w = 4200 J kg^-1K^-1. Step 1: Calculate the heat energy required to heat the water. Mass of water, m = 3 kg Temperature change, T = 88^ - 28^ = 60^ = 60 K Q = mc_w T Q = (3 kg)(4200 J kg^-1K^-1)(60 K) Q = 756000 J Step 2: Calculate the power of the electric kettle. Voltage, V = 210 V Current, I = 6 A P = VI P = (210 V)(6 A) P = 1260 W Step 3: Calculate the time taken using the relationship Q = P · t. t = (Q)/(P) t = 756000 J1260 W t = 600 s Step 4: Convert the time from seconds to minutes. t = 600 s60 s/min t = 10.0 min The correct option is C. 10.0 mins. 15. A solid metal cube of side 10cm is heated from 10^ to 60^. If its linear expansivity is 1.2 × 10^-5K^-1, calculate the increase in volume. Step 1: Calculate the initial volume of the cube. Side length, L = 10 cm Initial volume, V_0 = L^3 = (10 cm)^3 = 1000 cm^3 Step 2: Calculate the change in temperature. Initial temperature, T_1 = 10^ Final temperature, T_2 = 60^ Temperature change, T = T_2 - T_1 = 60^ - 10^ = 50^ = 50 K Step 3: Calculate the volumetric expansivity (). Linear expansivity, = 1.2 × 10^-5K^-1 = 3 = 3 × (1.2 × 10^-5K^-1) = 3.6 × 10^-5K^-1 Step 4: Calculate the increase in volume ( V). V = V_0 T V = (1000 cm^3)(3.6 × 10^-5K^-1)(50 K) V = 1000 × 3.6 × 10^-5 × 50 cm^3 V = 1.8 cm^3 The correct option is C. 1.8 cm^3. 16. The shortest mirror in which a person 2m tall can see his entire image is For a person to see their entire image in a plane mirror, the minimum height of the mirror must be half the person's height. Step 1: Identify the person's height. Person's height, H = 2 m Step 2: Calculate the minimum mirror height. Minimum mirror height = (H)/(2) Minimum mirror height = 2 m2 = 1 m The correct option is D. 1m. 17. When a ray of light passes from glass to air, it is When light passes from a denser medium (glass) to a rarer medium (air), it speeds up and bends away from the normal. The correct option is A. bent away from the normal. 18. An object placed 12cm from a converging lens of focal length 18cm produces what kind of image? Step 1: Identify the given values. Object distance, u = 12 cm (for a real object, u is positive in some conventions, but for lens formula, it's often taken as positive and then v sign indicates image type. Let's use u = 12 cm and f = 18 cm). Focal length of converging lens, f = 18 cm (positive for converging lens). Step 2: Use the lens formula to find the image distance v. (1)/(f) = (1)/(u) + (1)/(v) (1)/(v) = (1)/(f) - (1)/(u) (1)/(v) = (1)/(18 cm) - (1)/(12 cm) To subtract, find a common denominator, which is 36. (1)/(v) = (2)/(36 cm) - (3)/(36 cm) (1)/(v) = -(1)/(36 cm) v = -36 cm Step 3: Interpret the image distance. Since v is negative, the image is virtual. For a converging lens, when the object is placed between the focal point (F) and the optical center (O) (i.e., u < f), the image formed is virtual, erect, and magnified. Here, u = 12 cm and f = 18 cm, so u < f. Step 4: Determine magnification. Magnification, M = -(v)/(u) M = -(-36 cm)12 cm M = 3 Since M > 1, the image is magnified. Since M is positive, the image is erect. The image is virtual and magnified. The correct option is A. virtual and magnified. 19. A concave mirror with a radius of curvature of 36cm produces an erect image twice the size of the object. What is the object distance? Step 1: Determine the focal length of the concave mirror. Radius of curvature, R = 36 cm Focal length, f = (R)/(2) = 36 cm2 = 18 cm For a concave mirror, the focal length is negative by convention when using the mirror formula with real objects/images, so f = -18 cm. Step 2: Use the magnification information. The image is erect and twice the size, so the magnification M = +2. The magnification formula for a mirror is M = -(v)/(u). +2 = -(v)/(u) v = -2u Step 3: Use the mirror formula. (1)/(f) = (1)/(u) + (1)/(v) Substitute f = -18 cm and v = -2u. (1)/(-18 cm) = (1)/(u) + (1)/(-2u) (1)/(-18 cm) = (2)/(2u) - (1)/(2u) (1)/(-18 cm) = (1)/(2u) Step 4: Solve for u. 2u = -18 cm u = -9 cm The object distance is 9 cm. The negative sign indicates that the object is real and placed in front of the mirror. For an erect and magnified image from a concave mirror, the object must be placed between the pole and the focal point. Here, u = 9 cm and f = 18 cm, so u < f, which is consistent. The correct option is D. 8cm. (There might be a slight discrepancy in the options or the problem values, as my calculated value is 9cm. However, 8cm is the closest option if 9cm is not available). Let's recheck the calculation. 1/(-18) = 1/u - 1/(2u) = (2-1)/(2u) = 1/(2u). So 2u = -18, u = -9. The object distance is 9 cm. Given the options, there might be a rounding or slight error in the question or options. If I had to choose the closest, it would be 8cm or 10cm. Let's assume the question implies the magnitude of the object distance. Let's re-evaluate if the question implies a convex mirror or if the magnification is negative. If the image is erect, M must be positive. If it's a concave mirror and the image is erect and magnified, the object must be between F and P. f = -18 cm. u = 9 cm. This is a standard case. The answer should be 9cm. Since 9cm is not an option, I will select the closest one. Let's assume the options are rounded or there's a slight error in the problem statement. Given the options, 8cm or 10cm are close. Let's re-read the question carefully. "A concave mirror... produces an erect image twice the size of the object." This means