Selina solutions

Selina solutions

Selina solutions

Grade 7

CHAPTERS

Q11) A block of iron has dimensions 2m\times0.5m\times0.25m. The density of iron is 7.8\ g\ cm^{-3}. Find the mass of block.

Solution:

Length of the block of iron = 2m

Breadth of the block of iron = 0.5m

Height of the block of iron = 0.25 m

density = 7.8\ g\ cm^{-3}

= 7.8\times1000\ kg\ m^{-3}\ =\ 7800\ kg\ m^{-3}

Volume = l\times b\times h

= 2\times0.5\times0.25

= 0.25\ m^3

d = \frac{M}{V}

M = V\times d

= 0.25\times7800\ kg\ m^{-3}

= 1950 kg

Still have questions? Our expert teachers can help you out

Book a free class nowSolution:

Length of the block of iron = 2m

Breadth of the block of iron = 0.5m

Height of the block of iron = 0.25 m

density = 7.8\ g\ cm^{-3}

= 7.8\times1000\ kg\ m^{-3}\ =\ 7800\ kg\ m^{-3}

Volume = l\times b\times h

= 2\times0.5\times0.25

= 0.25\ m^3

d = \frac{M}{V}

M = V\times d

= 0.25\times7800\ kg\ m^{-3}

= 1950 kg

Still have questions? Our expert teachers can help you out

Book a free class nowLido

Courses

Race To Space

Quick Links

Terms & Policies

Terms & Policies