CBSE Class 7 Science Advanced Physics Measurements MCQs Set 04

Answer: (c) Joule (J)

Explanation:
1. Kelvin is the SI unit of temperature, erg is the CGS unit of energy and newton is the SI unit of force.
2. Energy and work are measured in the same unit, so the SI unit of energy is the joule (J).

Teacher's Note:
1. Learn the pairs: erg (CGS) and joule (SI) for energy; dyne (CGS) and newton (SI) for force.
2. 1 J = 10⁷ erg.

Answer: (d) 7

Explanation:
1. The curved path ABC is the perimeter given here, so \( \pi r = 22 \) km, that is \( \frac{22}{7} r = 22 \) km.
2. Radius r = 7 km, so the actual diameter = 2 × 7 = 14 km.
3. The scale is 1 cm = 2 km, so the diameter on the graph = 14 ÷ 2 = 7 cm.

Teacher's Note:
1. Use the perimeter of the arc shown, not an extra straight edge, because only the curve ABC is drawn.
2. Always divide the actual length by the scale value to get the length on paper.

Answer: (a) 54.85

Explanation:
1. The square runs from 4 to 12 on the X axis and from 3 to 11 on the Y axis, so each side = 8 divisions × 2 cm = 16 cm.
2. Area of the square = 16 × 16 = 256 cm².
3. The circle just fits inside the square, so its radius = 8 cm and its area = \( \frac{22}{7} \times 8 \times 8 \) = 201.14 cm².
4. Shaded area = 256 − 201.14 = 54.86 cm², which matches option (a) 54.85 (rounded).

Teacher's Note:
1. Count the divisions between the two marks on each axis and then multiply by 2 cm.
2. Shaded area means area of the square minus area of the circle.

Answer: (b) Circumference of the circle < Perimeter of the square

Explanation:
1. Each side of the square = 8 divisions × 2 cm = 16 cm, so its perimeter = 4 × 16 = 64 cm.
2. The circle has radius 8 cm, so its circumference = 2πr = \( 2 \times \frac{22}{7} \times 8 \) = 50.3 cm.
3. 50.3 cm is less than 64 cm, so the circumference of the circle is less than the perimeter of the square: (b).

Teacher's Note:
1. A circle inside a square always has a shorter boundary than the square.
2. Work out both numbers instead of guessing.

Answer: (d) All of these

Explanation:
1. Volume = \( \frac{1}{3} \times \frac{22}{7} \times 3.5 \times 3.5 \times 9 \) = 115.5 cm³.
2. Mass = density × volume = 8 × 115.5 = 924 g.
3. The weight of 924 g is 924 g wt = 924 gf.
4. In dynes: 924 g × 1000 cm s⁻² (since g = 10 m s⁻² = 1000 cm s⁻²) = 9,24,000 dynes.
5. All three values are the same weight, so the answer is (d) All of these.

Teacher's Note:
1. Gram weight and gram force are the same unit, and 1 gf = 1000 dynes when g = 10 m s⁻².
2. When one option says 'all of these', check each option before choosing it.

Answer: (b) Q > P > R

Explanation:
1. Least count of P = 100 cm ÷ 200 = 0.5 cm.
2. Least count of Q = 5 cm ÷ 20 = 0.25 cm.
3. Least count of R = 100 cm ÷ 100 = 1 cm.
4. The smaller the least count, the higher the accuracy. Q has the smallest and R the largest least count, so the order of accuracy is Q > P > R.

Teacher's Note:
1. Accuracy goes up when the least count goes down.
2. Find the least count of every scale before comparing.

Answer: (d) Both (a) and (b)

Explanation:
1. Mass of the liquid = 200 − 100 = 100 g.
2. Volume of the liquid = capacity of the bottle = 100 mL = 100 cm³.
3. Density = 100 g ÷ 100 cm³ = 1 g cm⁻³ = 1000 kg m⁻³.
4. Options (a) and (b) are the same value, so the answer is (d).

Teacher's Note:
1. 1 g cm⁻³ is the same as 1000 kg m⁻³; multiply by 1000 to change CGS density to SI.
2. Subtract the mass of the empty bottle first.

Answer: (a) Dynes cm⁻²

Explanation:
1. Pressure = force ÷ area. The CGS unit of force is dyne and of area is cm².
2. So the CGS unit of pressure is dyne cm⁻².
3. gf is a unit of force, erg is a unit of energy and g cm⁻³ is a unit of density.

Teacher's Note:
1. Write pressure as force per unit area and put in the CGS units.
2. In SI the matching unit is N m⁻².

Answer: (b) BCA

Explanation:
1. First note the given density (B): 5 g cm⁻³.
2. Then express 1 g as \( \frac{1}{1000} \) kg and 1 cm as \( \frac{1}{100} \) m (C).
3. Finally substitute these values and calculate: 5 × 10⁻³ kg ÷ 10⁻⁶ m³ = 5000 kg m⁻³ (A).
4. The order is B, C, A: (b) BCA.

Teacher's Note:
1. Every conversion starts with noting the given value.
2. Conversion factors are written down before they are substituted.

Answer: (c) 0.909 mm

Explanation:
1. The 22 threads cover 3 cm − 1 cm = 2 cm = 20 mm.
2. Distance between two successive threads = 20 mm ÷ 22 = 0.909 mm.
3. 0.909 mm = 0.0909 cm, so (c) 0.909 mm is correct.

Teacher's Note:
1. Divide the total length by the number of threads.
2. Check the unit: 0.0909 cm is the same as 0.909 mm, not 0.0909 mm.

Answer: (d) Both (b) and (c).

Explanation:
1. Option (a) is true: both are copper, so they have the same density and, with equal volumes, the same mass.
2. Option (b) is false: the triangulation method can be used to measure the width of a river.
3. Option (c) is false: relative density is the ratio of the density of a substance to the density of water; mass divided by volume is simply density.
4. So statements (b) and (c) are false: (d).

Teacher's Note:
1. Read the question again: it asks for the false statements.
2. Relative density has no unit because it is a ratio of two densities.

Answer: (b) 2500

Explanation:
1. 10 cm on the atlas = 1000 km, so 1 cm = 100 km.
2. 25 cm = 25 × 100 = 2500 km.

Teacher's Note:
1. First find what 1 cm stands for, then multiply.
2. Do not forget to write the unit of the answer.

Answer: (b) A → b; B → e; C → d; D → c; E → a

Explanation:
1. A density bottle is used to find the density of liquids: A → b.
2. When the temperature decreases, the density of a liquid increases: B → e.
3. The triangulation method measures the distance between the stars: C → d.
4. 10⁶ mg = 10⁶ × 10⁻³ g = 10³ g = 1 kg: D → c.
5. Velocity is measured in m s⁻¹: E → a.
6. The matching A → b; B → e; C → d; D → c; E → a is option (b).

Teacher's Note:
1. Convert the 10⁶ mg to kg step by step to avoid slips.
2. Match the sure pairs first and cancel the wrong options.

Answer: (d) Both (a) and (b)

Explanation:
1. Weight is a force, so it can be written in gravitational units.
2. In the CGS system the weight of a body is expressed in gram weight (g wt) or gram force (gf), which are the same unit.
3. g cm⁻³ is the unit of density. So the answer is (d).

Teacher's Note:
1. Gram weight and gram force are two names for the same unit.
2. In SI the matching units are newton or kilogram weight (kgf).

Answer: (b) 20

Explanation:
1. Distance AH = speed × time = 400 km h⁻¹ × 1 h = 400 km.
2. Scale: 1 cm = 20 km.
3. Distance on the graph = 400 ÷ 20 = 20 cm.

Teacher's Note:
1. The length of the river (920 km) is not needed; use the straight path of the glider.
2. Divide the real distance by the scale to get the length on paper.

Answer: (c) 37.6

Explanation:
1. The centre of the ring is at 6 on the Y axis. Radius of the circle through A = (11 − 6) × 2 cm = 10 cm.
2. Radius of the circle through B = (8 − 6) × 2 cm = 4 cm.
3. Circumference through A = 2π × 10 = 62.8 cm and through B = 2π × 4 = 25.1 cm.
4. Difference = 62.8 − 25.1 ≈ 37.7 cm, and the closest option is (c) 37.6.

Teacher's Note:
1. Count the divisions from the centre to the circle, not from the origin.
2. The exact value 2π × 6 is about 37.7 cm; the option given is the nearest one.

Answer: (d) \( \frac{1848}{7} \)

Explanation:
1. Outer radius = 10 cm and inner radius = 4 cm (found from the graph as 5 and 2 divisions of 2 cm).
2. Area of the big circle = \( \frac{22}{7} \times 10^2 = \frac{2200}{7} \) cm².
3. Area of the small circle = \( \frac{22}{7} \times 4^2 = \frac{352}{7} \) cm².
4. Area of the ring = \( \frac{2200}{7} - \frac{352}{7} = \frac{1848}{7} \) cm² (= 264 cm²): (d).

Teacher's Note:
1. Area of a ring = π(R² − r²).
2. Keep the fraction with 7 in the denominator while subtracting.

Answer: (a) 792 gf

Explanation:
1. Volume = \( \frac{4}{3} \pi r^3 = \frac{4}{3} \times \frac{22}{7} \times 27 \) = 113.14 cm³.
2. Mass = density × volume = 7 × 113.14 = 792 g.
3. The weight of a 792 g mass is 792 gf (= 7,92,000 dynes = 7.92 N), so the correct option is (a) 792 gf.

Teacher's Note:
1. Numbers in different options may look alike but belong to different units.
2. 792 N would be 79.2 kg and 792 kgf would be 792 kg, which are far too big.

Answer: (b) C > B > A

Explanation:
1. A scale divided into more parts has a smaller least count.
2. Least counts: A is the largest (10 parts), then B (50 parts), then C (100 parts).
3. The smaller the least count, the greater the accuracy, so the accuracy order is C > B > A.

Teacher's Note:
1. More divisions on the same length means a more accurate scale.
2. Read each inequality sign carefully before choosing.

Answer: (a) 2 g cm⁻³

Explanation:
1. Volume of the liquid = 200 mL = 200 cm³.
2. Density = 400 g ÷ 200 cm³ = 2 g cm⁻³ = 2000 kg m⁻³.
3. 1000 kg m⁻³ is only 1 g cm⁻³, so (b) is wrong.

Teacher's Note:
1. 1 mL is the same as 1 cm³.
2. Compare the units in the options before choosing.

Answer: (d) m s⁻²

Explanation:
1. Acceleration = change in velocity ÷ time = (m s⁻¹) ÷ s = m s⁻².
2. cm s⁻² is the CGS unit, so the SI unit is m s⁻².

Teacher's Note:
1. SI units use metre, kilogram and second.
2. m s⁻¹ is the unit of velocity, not acceleration.

Answer: (a) BACD

Explanation:
1. First read the three weights from the data (B): W₁ = 50 g, W₂ = 150 g, W₃ = 250 g.
2. Next write the formula for relative density (A).
3. Then substitute and calculate (C): relative density = \( \frac{250 - 50}{150 - 50} = 2 \).
4. Finally multiply by the density of water to get the density of the liquid (D): 2 × 1 = 2 g cm⁻³.
5. The sequence is B, A, C, D: (a) BACD.

Teacher's Note:
1. Data first, formula second, substitution third, final conversion last.
2. You cannot use the relative density in step D before it is calculated in step C.

Answer: (d) 1.33 mm

Explanation:
1. The pile of 15 coins stretches from the 0 mark to the 2 cm mark, so its length = 2 cm = 20 mm.
2. Thickness of one coin = 20 mm ÷ 15 = 1.33 mm.
3. 1.33 mm is option (d); 1.33 cm and 13.3 mm are ten times too large.

Teacher's Note:
1. Convert the length to millimetres before dividing, as the options are in mm and cm.
2. Read the scale from the zero mark carefully.

Answer: (d) Both (a) and (c).

Explanation:
1. Option (a) is true: the triangulation method is used for very distant objects such as stars.
2. Option (b) is false: the hollow cube has less mass for the same outer volume, so its overall density is less than that of the solid cube.
3. Option (c) is true: density = mass ÷ volume.
4. Statements (a) and (c) are true: (d).

Teacher's Note:
1. Compare the mass in the same volume to compare the density of two objects.
2. A hollow object has empty space inside, which lowers its density.

Answer: (a) 700

Explanation:
1. Scale: 1 cm on the atlas = 20 km on the ground.
2. 35 cm × 20 km cm⁻¹ = 700 km.

Teacher's Note:
1. Multiply the map length by the number of km one centimetre stands for.
2. A quick check: the answer must be larger than the 35 cm on paper.

Answer: (a) A → e; B → b; C → a; D → c; E → d

Explanation:
1. The relative density has the same numerical value as the density in the CGS system: A → e.
2. A physical balance measures mass in kg or g: B → b.
3. 10⁻⁶ kg = 10⁻⁶ × 10³ g = 10⁻³ g = 1 mg: C → a.
4. When the temperature increases, the density of gases decreases: D → c.
5. Large distances are measured by triangulation: E → d.
6. This is option (a).

Teacher's Note:
1. 10⁻⁶ kg = 10⁻³ g = 1 mg; use steps to be sure.
2. Check each option against the pairs you are sure of.

Answer: (b) 0.5 mm

Explanation:
1. Between the 0 mark and the 1 cm mark there are 20 equal divisions.
2. Value of one division = 1 cm ÷ 20 = 0.05 cm = 0.5 mm.
3. This least count is smaller than 1 mm, so the scale can measure 0.5 mm.

Teacher's Note:
1. The smallest measurement of a scale is its least count.
2. Count the spaces between the marks, not the marks themselves.

Answer: (d) Joule

Explanation:
1. Gram force (gf), kilogram force (kgf) and newton are units of force.
2. Joule is the unit of work and energy, so it is not a unit of force.

Teacher's Note:
1. Force: newton, dyne, gf, kgf. Energy: joule, erg.
2. Do not mix the units of force with the units of energy.

Answer: (d) All of these

Explanation:
1. 1 km = 1000 m.
2. 1 km = 1000 × 100 cm = 1,00,000 cm.
3. 1 km = 1,00,000 × 10 mm = 10,00,000 mm.
4. All three are correct, so the answer is (d) All of these.

Teacher's Note:
1. Multiply step by step: km to m (×1000), m to cm (×100), cm to mm (×10).
2. Check each option before choosing 'All of these'.

Answer: (b) \( \frac{11}{7} \) mm

Explanation:
1. The pile has 14 coins and it stretches from 0 to 22 on the scale.
2. Length of the pile = 22 divisions × 1 mm = 22 mm.
3. Thickness of one coin = \( \frac{22}{14} = \frac{11}{7} \) mm: (b).

Teacher's Note:
1. Count the coins in the pile carefully before dividing.
2. Cancel the common factor 2 in 22/14.

Answer: (a) \( \frac{4}{3} \)

Explanation:
1. From the graph, the volume at 100°C is 30 cm³ and at 150°C it is 40 cm³.
2. For the same mass, density = mass ÷ volume, so density is inversely proportional to volume.
3. \( \frac{d_{100}}{d_{150}} = \frac{V_{150}}{V_{100}} = \frac{40}{30} = \frac{4}{3} \): (a).

Teacher's Note:
1. Same mass means density changes opposite to volume.
2. Read the volume values for the two temperatures from the dashed lines.

Answer: (b) 2 g cm⁻³

Explanation:
1. Mass of the liquid = 1500 − 500 = 1000 g.
2. The liquid fills the container, so its volume = 500 cm³.
3. Density = 1000 ÷ 500 = 2 g cm⁻³.

Teacher's Note:
1. Subtract the mass of the empty container first.
2. Option (d) has the wrong unit, g m⁻³.

Answer: (c) 10 g cm⁻³

Explanation:
1. Volume = 10 × 5 × 2 = 100 cm³.
2. Density = 1000 g ÷ 100 cm³ = 10 g cm⁻³.

Teacher's Note:
1. Volume of a cuboid = length × breadth × height.
2. Check the unit of density in the option too.

Answer: (b) 1 : 5

Explanation:
1. Let each have mass m. Volume = mass ÷ density.
2. Volume of alloy = m ÷ 5 and volume of water = m ÷ 1 = m.
3. Ratio = \( \frac{m}{5} : m = 1 : 5 \): (b).

Teacher's Note:
1. For equal masses, the denser material takes less volume.
2. Write the volumes as m/density and cancel m.

Answer: (d) 1 g cm⁻³

Explanation:
1. Half of the body displaces 500 cm³, so the total volume of the body = 2 × 500 = 1000 cm³.
2. Weight 1000 gf means the mass is 1000 g.
3. Density = 1000 g ÷ 1000 cm³ = 1 g cm⁻³.

Teacher's Note:
1. Displaced volume equals the immersed volume, so double it to get the whole volume.
2. 1000 gf corresponds to a mass of 1000 g.

Answer: (c) 500 m

Explanation:
1. On the drawing AC = 6 cm, and the actual distance AC = 600 m, so the scale is 6 cm = 600 m, that is 1 cm = 100 m.
2. AB on the drawing = 5 cm, so the actual distance AB = 5 × 100 = 500 m.

Teacher's Note:
1. Find the scale from the side whose real length is known.
2. Then apply the same scale to the other side.

Answer: (c) A is true, but R is false.

Explanation:
1. A spring balance measures the pull of gravity on a body, so it measures weight. A is true.
2. The weight of a body becomes zero when no gravitational force acts on it, for example far away from all planets. R is false.
3. A is true but R is false: (c).

Teacher's Note:
1. Mass of a body can never be zero, but its weight can be.
2. Do not confuse the two quantities in assertion-reason questions.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. A body pushes aside a volume of liquid equal to the volume of its immersed part, whatever the liquid is. So R is true.
2. Both spheres are identical and immersed completely, so the immersed volumes are equal, and the displaced volumes of water and kerosene are equal. A is true.
3. R is the reason for A, so the answer is (a).

Teacher's Note:
1. Volume of displaced liquid depends on the body, not on the kind of liquid.
2. Density of the liquid matters only for floating and sinking.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Mass is the amount of matter in a body and it never becomes zero anywhere. A is true.
2. A common (physical) balance measures mass. R is true.
3. The reason a body's mass cannot be zero is not that a common balance measures it, so R does not explain A: (b).

Teacher's Note:
1. Both statements being true does not mean one explains the other.
2. Ask: does R tell why A happens? Here it only names the instrument.

Answer: (c) A is true, but R is false.

Explanation:
1. The 7 cm sphere has a larger volume than the 3 cm sphere, and a completely immersed solid displaces water equal to its own volume. So A is true.
2. R is stated for any body and says that the displaced water depends on the total volume. This is not true in general: a partly immersed body displaces water equal only to its immersed volume. So R is false.
3. A is true but R is false: (c).

Teacher's Note:
1. The displaced liquid depends on the immersed part of the body, which is the whole body only when it is fully inside the liquid.
2. Look out for words like 'any body' in the reason.