Exploring Magnets – Numerical Problems with Stepwise Solutions

CBSE Class 6 Science - Chapter 4: Exploring Magnets — 20 Numerical Problems with Stepwise Solutions

Content Bank — Important formulas & notes

This chapter is largely conceptual. Below are arithmetic relationships or useful concise notes that appear in numerical problems for Class 6 level.

Counting / rate of pick-up: If one magnet picks n nails per stroke, total after r strokes = n × r.
Series magnets: When magnets are placed end-to-end (N–S), their effective length adds; count as number of magnets × length of one for simple distance problems.
Percentage retained after demagnetisation (approx): Remaining magnetism = Initial × (1 − loss\%) (used for simple proportional problems, not rigorous physics).
Electromagnet turns: When given turns N and current I (basic class 6 level): more turns ⇒ stronger magnet (qualitative). For counting turns: total turns = sum of coils.
Field-line drawing fact: Field lines form closed loops. Use counting or proportion questions about number of lines started from a pole in diagrams.
Units reminder: Length in cm, m, counts (no unit) and time in seconds/minutes — keep consistent units in calculations.

20 Numerical Problems — Topic-wise (with Stepwise Solutions)

Each problem follows NCERT Class 6 concepts. Steps are simple arithmetic or logical steps to build problem-solving confidence.

Problem 1 (Picking nails — rate):

A strong bar magnet picks up 6 iron nails in one stroke. How many nails will it pick up in 7 similar strokes?

Solution:

Nails per stroke = 6.
Number of strokes = 7.
Total nails = 6 × 7 = 42.
Answer: 42 nails.

Problem 2 (Combining magnets length):

One small bar magnet is 4 cm long. If 5 such magnets are placed end-to-end (N–S, touching), what is the total length of the combined magnet?

Solution:

Length of one magnet = 4 cm.
Number of magnets = 5.
Total length = 4 cm × 5 = 20 cm.
Answer: 20 cm.

Problem 3 (Loss of pick-up when weakened):

A magnet could pick up 30 small pins before it was heated and weakened. After heating, its strength fell by 20%. How many pins can it now pick up?

Solution:

Initial capacity = 30 pins.
Loss = 20% of 30 = 0.20 × 30 = 6 pins lost.
Remaining capacity = 30 − 6 = 24 pins.
Answer: 24 pins.

Problem 4 (Magnetising by rubbing):

A student magnetises 12 steel pins by rubbing each pin 5 times with a magnet. If rubbing each pin takes 8 seconds per stroke, how much time did the student spend rubbing all pins (in minutes)?

Solution:

Number of pins = 12. Strokes per pin = 5. Total strokes = 12 × 5 = 60.
Time per stroke = 8 s. Total time in seconds = 60 × 8 = 480 s.
Convert to minutes: 480 s = 480/60 = 8 minutes.
Answer: 8 minutes.

Problem 5 (Induced magnetism count):

A long iron rod becomes magnetic when a magnet is brought close and it attracts 9 small nails. If two identical rods are placed side-by-side and both become equally magnetised, how many nails will both rods attract together?

Solution:

One rod attracts = 9 nails.
Two rods together attract = 9 × 2 = 18 nails.
Answer: 18 nails.

Problem 6 (Proportion of magnetised fraction):

A soft iron bar when held near a magnet becomes magnetic in 60% of its length. If the bar is 50 cm long, what length becomes magnetised?

Solution:

Total length = 50 cm. Magnetised fraction = 60% = 0.60.
Magnetised length = 0.60 × 50 = 30 cm.
Answer: 30 cm.

Problem 7 (Counting field lines):

In a simple diagram, 8 magnetic field lines are drawn coming out of the north pole of a magnet. If the same number of lines enter the south pole, what is the total number of field lines in the diagram (outside the magnet)?

Solution:

Lines leaving north pole = 8. Lines entering south pole = 8.
Total outside = 8 + 8 = 16 lines (note: lines form closed loops overall).
Answer: 16 lines outside the magnet.

Problem 8 (Compass deflection count):

A student places a strong magnet near a compass and observes 15° deflection from the north direction. If another magnet is placed on the opposite side producing an additional 10° deflection in the same direction, what is the total deflection observed?

Solution:

First deflection = 15°. Second deflection = 10° in same direction.
Total deflection = 15° + 10° = 25°.
Answer: 25° deflection.

Problem 9 (Field-line ratio):

A diagram shows 12 field lines emerging from a stronger magnet and 4 from a weaker magnet. What is the ratio of their strengths, based on the number of lines drawn?

Solution (qualitative arithmetic):

Stronger magnet lines = 12. Weaker = 4.
Ratio = 12 : 4 = 3 : 1.
Answer: 3 : 1 (stronger : weaker) based on field-line count in the diagram.

Problem 10 (Counting turns):

An electromagnet is made by winding 8 turns of insulated copper wire on a nail for the first layer and then 7 turns on top of it for a second layer. How many turns are there in total?

Solution:

First layer = 8 turns. Second layer = 7 turns.
Total turns = 8 + 7 = 15 turns.
Answer: 15 turns total.

Problem 11 (Proportional strength with turns - simple):

If an electromagnet with 10 turns picks up 20 paper clips, how many clips would be expected (simple proportional estimate) if the turns are increased to 15 turns (all else same)?

Solution (proportional estimate):

Initial turns = 10 ⇒ picks 20 clips.
Strength assumed proportional to turns: new clips = 20 × (15/10) = 20 × 1.5 = 30.
Answer (estimate): 30 paper clips.

Problem 12 (Average picks per trial):

In three trials, a magnet picked up 8, 10 and 9 nails respectively. What is the average number of nails picked per trial?

Solution:

Total nails = 8 + 10 + 9 = 27.
Number of trials = 3. Average = 27 / 3 = 9 nails per trial.
Answer: 9 nails (average).

Problem 13 (Difference in picks):

A magnet A picks 14 nails, magnet B picks 9 nails. How many more nails does A pick than B?

Solution:

Difference = 14 − 9 = 5.
Answer: Magnet A picks 5 more nails than B.

Problem 14 (Scaling counts):

If one magnet picks up 12 nails and you have 4 identical magnets used one after another, how many nails can be picked in total (assuming nails are unlimited)?

Solution:

Nails per magnet = 12. Number of magnets used sequentially = 4.
Total nails = 12 × 4 = 48.
Answer: 48 nails.

Problem 15 (Pairs of poles):

If you have 6 small bar magnets and you cut each magnet into two equal pieces (so each piece still has two poles), how many poles (north or south ends) are there in total after cutting?

Solution:

Each original magnet has 2 poles. Cutting does not create single poles; each piece still has 2 poles.
Number of pieces after cutting = 6 × 2 = 12 pieces. Each piece has 2 poles ⇒ total poles = 12 × 2 = 24.
Answer: 24 poles in total.

Problem 16 (Like-unlike interactions count):

Three magnets are placed such that their nearest poles facing each other are: N–S (pair 1), N–N (pair 2), S–S (pair 3). How many of these pairs will attract each other?

Solution:

Attraction occurs for unlike poles (N–S). Repulsion for like (N–N or S–S).
Pair 1 (N–S) attracts. Pair 2 (N–N) repels. Pair 3 (S–S) repels.
Number attracting = 1.
Answer: 1 pair attracts.

Problem 17 (Length conversion for magnet):

A magnet is 0.25 m long. Express its length in centimetres (cm).

Solution:

1 m = 100 cm. So 0.25 m = 0.25 × 100 = 25 cm.
Answer: 25 cm.

Problem 18 (Time for experiment):

An experiment requires moving a magnet over a strip of iron 40 times. If each movement takes 3 seconds, how long will the experiment take in minutes?

Solution:

Total moves = 40. Time per move = 3 s. Total time = 40 × 3 = 120 s.
Convert to minutes: 120 / 60 = 2 minutes.
Answer: 2 minutes.

Problem 19 (Percentage retained after gentle hammering):

A magnet had strength measured by ability to pick 50 pins. After gentle hammering, it can pick only 40 pins. What percentage of its original strength does it retain (rounded to whole %)?

Solution:

Original = 50 pins. After = 40 pins.
Fraction retained = 40 / 50 = 0.8 ⇒ percentage = 0.8 × 100 = 80%.
Answer: 80% retained.

Problem 20 (Dividing magnetised objects):

A student divides a long magnetic rod into 4 equal segments. If the total number of poles on the rod is counted as 8 before cutting (since each pole counts separately on the length), how many poles will each segment have after cutting (assuming poles distribute equally)?

Solution:

Total poles = 8. Number of segments = 4.
Poles per segment = 8 / 4 = 2.
Answer: Each segment will have 2 poles.

Exploring Magnets – Numerical Problems with Stepwise Solutions

CBSE Class 6 Science — Chapter 4: Exploring Magnets

Content Bank — Important formulas & notes

20 Numerical Problems — Topic-wise (with Stepwise Solutions)

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