Fill
in the blanks
(a)
The volume of a cube of side 1 cm is equal to.....m3
(b)
The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is
equal to ... (mm)2
(c) A vehicle moving with a speed of 18 km h–1covers....m
in 1 s
(d)
The relative density of lead is 11.3. Its density is ....g cm–3or .
...kg m–3.
Solution
Solution
(a)
Length of edge = 1cm = 1/100 m
Volume of the cube = side3
Plug the value of
side, we get
Volume of the cube =
(1/100 m)3
Answer: volume of a cube of side
1 cm is equal to 10–6 m3.
(b)
Given that
Radius, r = 2.0 cm = 20 mm
(convert cm to mm)
Height, h = 10.0 cm =100 mm
The formula of total surface area
of a cylinder S = 2πr (r + h)
Plug the values in this formula,
we get
Surface area of a cylinder S = 2πr (r + h = 2 × 3.14×20 (20+100)
= 15072 =
1.5 × 104 mm2
Answer
The surface area of a solid cylinder of radius
2.0 cm and height 10.0 cm is equal to 1.5 × 104 (mm)2
(c)
Using the conversion,
Given that
Time, t = 1 sec
speed = 18 km h-1 = 18 km / hour
1 km = 1000 m and
1hour = 3600 sec
Speed = 18 × 1000
/3600 sec = 5 m /sec
Use formula
Speed = distance / time
Cross multiply it, we
get
Distance = Speed × Time = 5 × 1 =
5 m
Answer: A vehicle moving with a speed of 18 km h–1covers
5 m in 1 s
(d)
Formula
Density of lead = Relative
density of lead × Density
of water
Density of water = 1 g/cm3
Plug the values, we
get
Density of lead = 11.3 × 1 g/ cm 3
=
11.3 g cm -3
1 cm = (1/100 m) =10–2
m3
1 g = 1/1000 kg = 10-3 kg
Density of lead = 11.3
g cm-3 = 11.3
Plug the value of 1 cm
and 1 gram
11.3 g/cm3 = 11.3 × 10-3
kg (10-2m)-3 = 11.3 ×10– 3 × 106
kg m-3 =1.13 × 103 kg m –3
Answer:
The relative density of lead is
11.3. Its density is 11.3 g cm -3.g cm–3 or 1.13 × 103 kg m–3.
Nice
ReplyDeleteThanks a lot !!
ReplyDeleteThx for your help
ReplyDeleteThnxx
ReplyDeleteVery helpful
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ReplyDeleteChal bakwas
ReplyDeleteDensity of lead = Relative density of lead × Density of water
DeleteDensity of water = 1 g/cm3
Plug the values, we get
Density of lead = 11.3 × 1 g/ cm 3
= 11.3 g cm -3
1 cm = (1/100 m) =10–2 m3
1 g = 1/1000 kg = 10-3 kg
Density of lead = 11.3 g cm-3 = 11.3
Plug the value of 1 cm and 1 gram
11.3 g/cm3 = 11.3 × 10-3 kg (10-2m)-3 = 11.3 ×10– 3 × 106 kg m-3 =1.13 × 103 kg m –3
Answer:
The relative density of lead is 11.3. Its density is 11.3 g cm -3.g cm–3 or 1.13 × 103 kg m–3.
DeleteHome class 9 class 10 Class 11 Class 12 CBSE syllabus CTET English Spoken Sample paper Question papers ▼
Tuesday, 21 May 2013
Fill in the blanks (a) The volume of a cube of side 1 cm is equal to.....m3 (b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to ... (mm)2
Fill in the blanks
(a) The volume of a cube of side 1 cm is equal to.....m3
(b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to ... (mm)2
(c) A vehicle moving with a speed of 18 km h–1covers....m in 1 s
(d) The relative density of lead is 11.3. Its density is ....g cm–3or . ...kg m–3.
Solution
(a)
Length of edge = 1cm = 1/100 m
Volume of the cube = side3
Plug the value of side, we get
Volume of the cube = (1/100 m)3
Answer: volume of a cube of side 1 cm is equal to 10–6m3.
(b)
Given that
Radius, r = 2.0 cm = 20 mm (convert cm to mm)
Height, h = 10.0 cm =100 mm
The formula of total surface area of a cylinder S = 2πr (r + h)
Plug the values in this formula, we get
Surface area of a cylinder S = 2πr (r + h = 2 × 3.14×20 (20+100)
= 15072 = 1.5 × 104 mm2
Answer
The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to 1.5 × 104 (mm)2
(c)
Using the conversion,
Given that
Time, t = 1 sec
speed = 18 km h-1 = 18 km / hour
1 km = 1000 m and 1hour = 3600 sec
Speed = 18 × 1000 /3600 sec = 5 m /sec
Use formula
Speed = distance / time
Cross multiply it, we get
Distance = Speed × Time = 5 × 1 = 5 m
Answer: A vehicle moving with a speed of 18 km h–1covers 5 m in 1 s
(d)
Formula
Density of lead = Relative density of lead × Density of water
Density of water = 1 g/cm3
Plug the values, we get
Density of lead = 11.3 × 1 g/ cm 3
= 11.3 g cm -3
1 cm = (1/100 m) =10–2 m3
1 g = 1/1000 kg = 10-3 kg
Density of lead = 11.3 g cm-3 = 11.3
Plug the value of 1 cm and 1 gram
11.3 g/cm3 = 11.3 × 10-3 kg (10-2m)-3 = 11.3 ×10– 3 × 106 kg m-3 =1.13 × 103 kg m –3
Answer:
The relative density of lead is 11.3. Its density is 11.3 g cm -3.g cm–3 or 1.13 × 103 kg m–3.
Unknown at 15:26
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12 comments:
Unknown7 September 2017 at 12:15
Nice
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palwasha khan19 September 2017 at 03:53
Thanks a lot !!
Reply
Unknown30 June 2018 at 09:14
Thx for your help
Reply
Unknown5 July 2018 at 08:52
Thnxx
Reply
Unknown8 July 2018 at 18:08
Very helpful
Reply
Unknown10 July 2018 at 10:58
Thnkzzzz
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Unknown18 July 2018 at 07:09
Nice
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Unknown10 September 2018 at 08:30
Chal bakwas
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Replies
Unknown26 April 2019 at 21:02
Density of lead = Relative density of lead × Density of water
Density of water = 1 g/cm3
Plug the values, we get
Density of lead = 11.3 × 1 g/ cm 3
= 11.3 g cm -3
1 cm = (1/100 m) =10–2 m3
1 g = 1/1000 kg = 10-3 kg
Density of lead = 11.3 g cm-3 = 11.3
Plug the value of 1 cm and 1 gram
11.3 g/cm3 = 11.3 × 10-3 kg (10-2m)-3 = 11.3 ×10– 3 × 106 kg m-3 =1.13 × 103 kg m –3
Answer:
The relative density of lead is 11.3. Its density is 11.3 g cm -3.g cm–3 or 1.13 × 103 kg m–3.
Reply
Niloy Das10 October 2018 at 21:27
this very helpful
Reply
Unknown13 February 2019 at 16:08
A Side 1 ko
Side 3 ka answer
Reply
Unknown26 April 2019 at 21:01
Thnx fot giving the answer
Reply
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this very helpful
ReplyDeleteA Side 1 ko
ReplyDeleteSide 3 ka answer
Thnx fot giving the answer
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ReplyDeleteThanks I clear my doubts
ReplyDeleteI like you solution... Thanks a lot...
ReplyDeleteI like you solution... Thanks a lot...
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