CBSE Class 7 Science Advanced Physics Measurements MCQs Set 02

Answer: (d) 1400

Explanation:
1. Volume of the stone = mass ÷ density = 2 ÷ (5 × 10³) = 4 × 10⁻⁴ m³.
2. Since 1 m³ = 10⁶ cm³, the volume of the stone = 4 × 10⁻⁴ × 10⁶ = 400 cm³.
3. The stone pushes the liquid level up by its own volume, so the new reading = 1000 + 400 = 1400 cm³.

Teacher's Note:
1. The rise in the reading of a graduated cylinder equals the volume of the immersed solid.
2. Convert m³ to cm³ before adding to the reading, because the cylinder reads in cm³.

Answer: (d) 10⁶

Explanation:
1. Mass = weight ÷ g = 2512 ÷ 10 = 251.2 kg.
2. Radius = 4 cm ÷ 2 = 2 cm = 0.02 m and height = 20 cm = 0.2 m.
3. Volume = πr²h = 3.14 × (0.02)² × 0.2 = 2.512 × 10⁻⁴ m³.
4. Density = mass ÷ volume = 251.2 ÷ (2.512 × 10⁻⁴) = 10⁶ kg m⁻³.

Teacher's Note:
1. The figure gives the diameter (4 cm), so halve it before using πr²h.
2. Change every length to metres to get the density directly in kg m⁻³.

Answer: (b) 100

Explanation:
1. Density of the liquid = 3000 kg m⁻³ = 3 g cm⁻³, so the liquid is 3 times as heavy as water for the same volume.
2. Let the empty bottle have mass m₁. Mass of liquid = 400 − m₁ and mass of water = 200 − m₁.
3. 400 − m₁ = 3 × (200 − m₁) = 600 − 3m₁, so 2m₁ = 200 and m₁ = 100 g.

Teacher's Note:
1. A density bottle holds the same volume of every liquid, so the masses of liquid and water are in the ratio of their densities.
2. Check: liquid 300 g, water 100 g, ratio 3 : 1.

Answer: (d) 10⁻⁹

Explanation:
1. 1 g = 10⁻³ kg and 1 m³ = (100 cm)³ = 10⁶ cm³.
2. So 1 g m⁻³ = 10⁻³ kg ÷ 10⁶ cm³ = 10⁻³ × 10⁻⁶ kg cm⁻³ = 10⁻⁹ kg cm⁻³.

Teacher's Note:
1. The mass unit gives a factor 10⁻³ and the volume unit gives a factor 10⁻⁶.
2. Multiply the two factors: 10⁻³ × 10⁻⁶ = 10⁻⁹.

Answer: (b) A = B < C

Explanation:
1. Least count of A = 1 m ÷ 500 = 0.002 m = 2 mm.
2. Least count of B = 1 cm ÷ 5 = 0.2 cm = 2 mm.
3. Least count of C = 1 mm ÷ 2 = 0.5 mm.
4. The smaller the least count, the more accurate the scale. A and B have the same least count, and C has a smaller one, so A = B < C in accuracy.

Teacher's Note:
1. Accuracy goes up as the least count goes down.
2. Write all three least counts in the same unit (mm) before comparing.

Answer: (d) 1.45 mm

Explanation:
1. The coil starts at 1.2 cm and ends at 4.1 cm on the scale, so its length = 4.1 − 1.2 = 2.9 cm.
2. The number of turns of the wire in the coil = 20.
3. Diameter of the wire = 2.9 cm ÷ 20 = 0.145 cm = 1.45 mm.

Teacher's Note:
1. Wind the wire closely, so that each turn touches the next one; then length of coil ÷ number of turns = thickness of wire.
2. Always count the turns, not the lines at the ends.

Answer: (d) 1/8

Explanation:
1. Volume of a sphere = \( \frac{4}{3}\pi r^3 \), so doubling the radius makes the volume 2³ = 8 times.
2. The mass stays the same, and density = mass ÷ volume.
3. New density = m ÷ 8V = \( \frac{1}{8} \) of the original density, so the answer is 1/8.

Teacher's Note:
1. Volume depends on the cube of the radius, so the effect is 2³ = 8.
2. Same mass in 8 times the space means one-eighth of the density.

Answer: (c) 312

Explanation:
1. The paper cutting is a big rectangle of 11 divisions (from 5 to 16) × 12 divisions with a rectangle of 6 divisions (from 8 to 14) × 9 divisions removed.
2. Area in divisions² = 11 × 12 − 6 × 9 = 132 − 54 = 78 (the same as 9 × 3 + 11 × 3 + 9 × 2 = 78).
3. One division = 2 cm, so 1 division² = 4 cm². Area = 78 × 4 = 312 cm².

Teacher's Note:
1. Split an odd shape into rectangles, or subtract a cut-out from a big rectangle.
2. Remember to multiply by the square of the scale (2 × 2 = 4).

Answer: (b) 2000

Explanation:
1. Mass of water = 100 − 50 = 50 g, so the volume of the bottle = 50 cm³ (density of water 1 g cm⁻³).
2. Mass of liquid = 150 − 50 = 100 g in the same 50 cm³.
3. Density of the liquid = 100 ÷ 50 = 2 g cm⁻³ = 2000 kg m⁻³.

Teacher's Note:
1. The water reading gives the volume of the bottle.
2. Multiply g cm⁻³ by 1000 to get kg m⁻³.

Answer: (a) 157

Explanation:
1. Radius = 20 cm ÷ 2 = 10 cm = 0.1 m and height = 10 cm = 0.1 m.
2. Volume = πr²h = 3.14 × (0.1)² × 0.1 = 0.00314 m³.
3. Mass = density × volume = 5 × 10³ × 0.00314 = 15.7 kg.
4. Weight = mg = 15.7 × 10 = 157 N.

Teacher's Note:
1. The figure shows the diameter, so find the radius first.
2. Weight = mass × g, and the answer is in newton because the mass is in kg.

Answer: (b) 2 × 10⁴

Explanation:
1. Volume of the alloy = 20 − 10 = 10 cm³ = 10 × 10⁻⁶ m³ = 10⁻⁵ m³.
2. Mass = 200 g = 0.2 kg.
3. Density = 0.2 ÷ 10⁻⁵ = 2 × 10⁴ kg m⁻³.

Teacher's Note:
1. Volume of the solid = final reading − initial reading.
2. Quick check: 200 g ÷ 10 cm³ = 20 g cm⁻³ = 20000 kg m⁻³.

Answer: (c) 1.8 mm

Explanation:
1. The pile of 10 coins starts at 0.6 cm and ends at 2.4 cm on the scale, so its height = 2.4 − 0.6 = 1.8 cm = 18 mm.
2. Thickness of one coin = 18 mm ÷ 10 = 1.8 mm.

Teacher's Note:
1. Measuring many coins together gives a more accurate thickness than measuring one coin.
2. Do not start reading from the zero mark if the pile does not start there.

Answer: (a) 2

Explanation:
1. Mass is fixed, so density becomes \( \frac{1}{8} \) when the volume becomes 8 times.
2. \( \frac{4}{3}\pi r_1^3 = 8 \times \frac{4}{3}\pi r^3 \), so r₁³ = 8r³ and r₁ = 2r.

Teacher's Note:
1. Volume goes as r³, so an 8-fold volume needs a 2-fold radius (since 2³ = 8).
2. Do not mix this with the 'radius doubled' question, where the density falls to 1/8.

Answer: (c) 93

Explanation:
1. Split the cross into three rectangles: the vertical bar from x = 8 to 12 and y = 2 to 17 is 4 × 15 = 60 cm².
2. The left arm from x = 3 to 8 and y = 8 to 11 is 5 × 3 = 15 cm², and the right arm from x = 12 to 18 is 6 × 3 = 18 cm².
3. Total area = 60 + 15 + 18 = 93 cm².

Teacher's Note:
1. Avoid counting the central square twice by dividing the shape into non-overlapping rectangles.
2. Here each division is 1 cm, so no scale factor is needed.

Answer: (c) 312

Explanation:
1. The paper cutting is a big rectangle of 11 divisions (from 5 to 16) × 12 divisions with a rectangle of 6 divisions (from 8 to 14) × 9 divisions removed.
2. Area in divisions² = 11 × 12 − 6 × 9 = 132 − 54 = 78 (the same as 9 × 3 + 11 × 3 + 9 × 2 = 78).
3. One division = 2 cm, so 1 division² = 4 cm². Area = 78 × 4 = 312 cm².

Teacher's Note:
1. Different splits of the shape give the same total, so use the one that is easiest to count.
2. Check that the scale factor is applied to the area, not to the number of divisions.

Answer: (d) 1.45 mm

Explanation:
1. The coil starts at 1.2 cm and ends at 4.1 cm on the scale, so its length = 4.1 − 1.2 = 2.9 cm.
2. The number of turns of the wire in the coil = 20.
3. Diameter of the wire = 2.9 cm ÷ 20 = 0.145 cm = 1.45 mm.

Teacher's Note:
1. Winding the wire many times turns a tiny length into one that a scale can read.
2. If the turns are loose, the diameter will come out too large.

Answer: (a) A → b, B → a, C → d, D → c

Explanation:
1. A measuring cylinder measures volume of liquids in ml or cc, so A → b.
2. kgf and gf are the gravitational units of weight, so B → a.
3. 1 g cm⁻³ = 1000 kg m⁻³, so C → d.
4. With no gravitational force, a body has zero weight, so D → c.
5. The correct set is A → b, B → a, C → d, D → c.

Teacher's Note:
1. Match the pairs you are sure about first, then eliminate options.
2. kgf and gf are called gravitational units because they depend on the pull of the earth.

Answer: (a) b a c d

Explanation:
1. First note the given data: d_A : d_B = 15 : 2 and equal masses (b).
2. Then write d = m ÷ V for each liquid, V_A = m ÷ d_A and V_B = m ÷ d_B (a).
3. Divide the two equations to get V_A ÷ V_B = d_B ÷ d_A (c).
4. Finally substitute: V_A ÷ V_B = 2 ÷ 15 (d). The order is b a c d.

Teacher's Note:
1. Collect the given data, write the formula, simplify, then substitute numbers.
2. Substituting numbers is always the last step.

Answer: (d) a d b c

Explanation:
1. Convert each unit into g cm⁻³: 1 kg m⁻³ = 10⁻³ g cm⁻³ (a), 1 g m⁻³ = 10⁻⁶ g cm⁻³ (d) and 1 kg cm⁻³ = 1000 g cm⁻³ (b).
2. Only after all the conversions are done can the values be compared and arranged in ascending order (c).
3. So every conversion (a, d, b) comes before the comparison (c). The sequence is a d b c.

Teacher's Note:
1. The comparison step needs all the converted values, so it must be last.
2. Ascending order of the units here is g m⁻³, kg m⁻³, g cm⁻³, kg cm⁻³.

Answer: (c) The mass of a body can be zero.

Explanation:
1. Weight = mg, so the weight is zero where g = 0 (no gravitational force). Statement (a) is correct.
2. Where g is not zero, the weight is greater than zero. Statement (b) is correct.
3. Mass is the amount of matter in a body, and a body always has some matter, so mass can never be zero. Statement (c) is incorrect.
4. Option (d) includes (a), which is correct, so (d) is wrong.

Teacher's Note:
1. Weight can vanish, mass cannot.
2. Read option (d) carefully: it is wrong because one of its two parts is a true statement.

Answer: (b) 2000

Explanation:
1. Mass of water = 100 − 50 = 50 g, so the volume of the bottle = 50 cm³ (density of water 1 g cm⁻³).
2. Mass of liquid = 150 − 50 = 100 g in the same 50 cm³.
3. Density of the liquid = 100 ÷ 50 = 2 g cm⁻³ = 2000 kg m⁻³.

Teacher's Note:
1. Subtract the empty-bottle mass from each filled reading before doing anything else.
2. A quick check: the liquid is twice as heavy as water for the same volume.

Answer: (d) 2 mm

Explanation:
1. The least count = 1 m ÷ 500 = 0.002 m.
2. 0.002 m = 2 mm, so the smallest measurement is 2 mm.

Teacher's Note:
1. The smallest measurement of a scale is its least count = length of scale ÷ number of divisions.
2. Convert metres to millimetres by multiplying by 1000.

Answer: (d) All the above.

Explanation:
1. Pressure = force ÷ area.
2. N m⁻² (SI), gwt cm⁻² (force in gram weight, area in cm²) and kgwt m⁻² (force in kilogram weight, area in m²) are all force per unit area.
3. So all the above are units of pressure.

Teacher's Note:
1. Any force unit divided by any area unit is a unit of pressure.
2. N m⁻² is the SI unit and is called pascal.

Answer: (b) A = B < C

Explanation:
1. Least count of A = 1 m ÷ 500 = 0.002 m = 2 mm.
2. Least count of B = 1 cm ÷ 5 = 0.2 cm = 2 mm.
3. Least count of C = 1 mm ÷ 2 = 0.5 mm.
4. The smaller the least count, the more accurate the scale. A and B have the same least count, and C has a smaller one, so A = B < C in accuracy.

Teacher's Note:
1. Compare scales by finding the least count of each in one common unit.
2. A scale that can read smaller lengths is the more accurate scale.

Answer: (d) 1/8

Explanation:
1. Doubling the radius makes the volume 2³ = 8 times.
2. With the same mass, density = m ÷ 8V = \( \frac{1}{8} \) of the original, so the answer is 1/8.

Teacher's Note:
1. Volume is proportional to r³.
2. More volume for the same mass means lower density.

Answer: (a) 157

Explanation:
1. Radius = 20 cm ÷ 2 = 10 cm = 0.1 m and height = 10 cm = 0.1 m.
2. Volume = πr²h = 3.14 × (0.1)² × 0.1 = 0.00314 m³.
3. Mass = density × volume = 5 × 10³ × 0.00314 = 15.7 kg.
4. Weight = mg = 15.7 × 10 = 157 N.

Teacher's Note:
1. Find the volume of the cylinder, then the mass, then the weight, in that order.
2. Do not forget to change cm to m before using the density in kg m⁻³.

Answer: (c) 350 km

Explanation:
1. On the atlas, the distance between A and B is 7 cm and the scale is 1 cm = 50 km.
2. Actual distance = 7 × 50 = 350 km.
3. 35 km and 3500 m are not equal to 350 km, so (a), (b) and (d) are wrong.

Teacher's Note:
1. Map distance × scale gives the real distance.
2. The scale must be given or read from the map; here it is 1 cm = 50 km.

Answer: (c) 93

Explanation:
1. Split the cross into three rectangles: the vertical bar from x = 8 to 12 and y = 2 to 17 is 4 × 15 = 60 cm².
2. The left arm from x = 3 to 8 and y = 8 to 11 is 5 × 3 = 15 cm², and the right arm from x = 12 to 18 is 6 × 3 = 18 cm².
3. Total area = 60 + 15 + 18 = 93 cm².

Teacher's Note:
1. Break the cross into a vertical bar and two arms to avoid counting the middle twice.
2. Add the three areas: 60 + 15 + 18.

Answer: (c) 1.8 mm

Explanation:
1. The pile of 10 coins starts at 0.6 cm and ends at 2.4 cm on the scale, so its height = 2.4 − 0.6 = 1.8 cm = 18 mm.
2. Thickness of one coin = 18 mm ÷ 10 = 1.8 mm.

Teacher's Note:
1. Divide the length of the stack by the number of coins to get one thickness.
2. 18 mm ÷ 10 = 1.8 mm, so look for the option in millimetres.

Answer: (c) A → d, B → a, C → b, D → c

Explanation:
1. A completely immersed solid displaces liquid equal to its own volume, so A → d.
2. 1 quintal = 100 kg = 100 × 1000 g = 10⁵ g, so B → a.
3. 1 tonne = 1000 kg = 1000 × 1000 g = 10⁶ g = 10⁶ × 1000 mg = 10⁹ mg, so C → b.
4. kgwt m⁻² is force per unit area, so D → c.
5. The correct set is A → d, B → a, C → b, D → c.

Teacher's Note:
1. Convert the mass units step by step: tonne to kg, kg to g, g to mg.
2. A force unit divided by an area unit is always a pressure unit.

Answer: (c) b a c d

Explanation:
1. First the objects are immersed in the liquids (b).
2. The liquid that overflows is collected and measured (a).
3. The same experiment is repeated at different places and times (c).
4. Finally the conclusion is stated (d). The order is b a c d.

Teacher's Note:
1. This follows the steps of science: observation, measurement, systematization and inference.
2. The conclusion always comes last.

Answer: (b) a b d c

Explanation:
1. Write the conversions 1 kg = 1000 g and 1 m = 100 cm (a).
2. Substitute them: 1 newton = 1000 g × 100 cm s⁻² = 10⁵ g cm s⁻² (b).
3. Replace 1 g cm s⁻² by 1 dyne (d).
4. Finally state the relation 1 newton = 10⁵ dyne (c). The order is a b d c.

Teacher's Note:
1. Conversions first, then substitution, then renaming the unit, then the final statement.
2. Remember: 1 N = 10⁵ dyne.

Answer: (c) The density of a body can be measured by using a spring balance.

Explanation:
1. A beam balance compares masses, so statement (a) is true, and it does not measure weight, so statement (d) is true.
2. A spring balance measures the pull of the earth, that is, weight, so statement (b) is true.
3. A spring balance alone gives only the weight. Density needs both mass and volume, so a spring balance by itself cannot measure density, and statement (c) is wrong.

Teacher's Note:
1. Density = mass ÷ volume, so an instrument that gives only weight cannot give it directly.
2. A density bottle with a physical balance is used for liquids.

Answer: (b) 100

Explanation:
1. Density of the liquid = 3000 kg m⁻³ = 3 g cm⁻³, so the liquid is 3 times as heavy as water for the same volume.
2. Let the empty bottle have mass m₁. Mass of liquid = 400 − m₁ and mass of water = 200 − m₁.
3. 400 − m₁ = 3 × (200 − m₁) = 600 − 3m₁, so 2m₁ = 200 and m₁ = 100 g.

Teacher's Note:
1. Use the fact that the same bottle holds the same volume of water and of liquid.
2. Convert the density 3000 kg m⁻³ to 3 g cm⁻³ before using the masses in grams.

Answer: (c) 10000

Explanation:
1. Least count = 0.1 mm = 0.1 × 10⁻³ m = 10⁻⁴ m.
2. Number of parts = 1 m ÷ 10⁻⁴ m = 10⁴ = 10000.

Teacher's Note:
1. Number of divisions = total length ÷ least count.
2. Convert the least count to metres first.

Answer: (a) joule

Explanation:
1. The joule (J) is the SI unit of energy.
2. The watt is the unit of power and the newton is the unit of force, so (b), (c) and (d) are wrong. The answer is joule.

Teacher's Note:
1. Energy: joule (SI) and erg (CGS).
2. Do not confuse watt (joule per second) with joule.

Answer: (c) One mm in a scale is divided into 2 equal parts.

Explanation:
1. Least count of (a) = 1 m ÷ 1000 = 1 mm.
2. Least count of (b) = 1 cm ÷ 2 = 5 mm.
3. Least count of (c) = 1 mm ÷ 2 = 0.5 mm.
4. Least count of (d) = 1 cm ÷ 5 = 2 mm.
5. The smallest least count is 0.5 mm, so the scale in (c) is the most accurate.

Teacher's Note:
1. Convert every least count to the same unit.
2. Smaller least count means more accurate reading.

Answer: (a) 2

Explanation:
1. Mass is fixed, so density becomes \( \frac{1}{8} \) when the volume becomes 8 times.
2. \( \frac{4}{3}\pi r_1^3 = 8 \times \frac{4}{3}\pi r^3 \), so r₁³ = 8r³ and r₁ = 2r.

Teacher's Note:
1. Density and volume are inversely related when the mass is fixed.
2. An 8 times larger volume needs a radius that is 2 times larger.

Answer: (d) 10⁶

Explanation:
1. Mass = weight ÷ g = 2512 ÷ 10 = 251.2 kg.
2. Radius = 4 cm ÷ 2 = 2 cm = 0.02 m and height = 20 cm = 0.2 m.
3. Volume = πr²h = 3.14 × (0.02)² × 0.2 = 2.512 × 10⁻⁴ m³.
4. Density = mass ÷ volume = 251.2 ÷ (2.512 × 10⁻⁴) = 10⁶ kg m⁻³.

Teacher's Note:
1. Always check that weight is converted into mass using g.
2. The answer 10⁶ kg m⁻³ comes straight from the numbers in the question.

Answer: (a) 10

Explanation:
1. Scale: 100 cm on the atlas stands for 15,000 km, so 1 cm = 15,000 ÷ 100 = 150 km.
2. Distance on the atlas = 1500 ÷ 150 = 10 cm.

Teacher's Note:
1. Find the real distance for 1 cm first, then divide.
2. Real distance ÷ scale gives the map distance.