CBSE Class 10 Answered
The adjoining figure shows the cross-section of a railway tunnel.
The radius of the tunnel is 3.5m (i.e., OA=3.5m) and ∠AOB=90o. Calculate :
i. the height of the tunnel.
ii. the perimeter of its cross section, including base.
iii. the area of the cross-section
iv. the internal surface area of the tunnel, excluding base, if its length is 50m.
(Ans: (i) 5.97m (ii) 21.44m (iii) 28.875 sq m (iv) 825 sq m)
The radius of the tunnel is 3.5m (i.e., OA=3.5m) and ∠AOB=90o. Calculate :
i. the height of the tunnel.
ii. the perimeter of its cross section, including base.
iii. the area of the cross-section
iv. the internal surface area of the tunnel, excluding base, if its length is 50m.
(Ans: (i) 5.97m (ii) 21.44m (iii) 28.875 sq m (iv) 825 sq m)
Asked by rushabhjain.av | 30 Jan, 2019, 09:28: PM
Expert Answer
Height of the tunnel :
Height of the tunnel = OC+OM
ΔAOB is right angled triangle and OA = OB = 3.5 m
Since OM is perpendicular drawn from O and ΔAOB is isoceless, we have OM = AM = h
also h = OA/√2 = 3.5 / √2 ≈ 2.47 m
Hence height of tunnel = 3.5+2.47 = 5.97 m
--------------------------------------------------------
Perimeter of cross section including its base:
since 
AOB is 90°, partial circular part of cross section = (3/4)th of full circular cross section.
AOB is 90°, partial circular part of cross section = (3/4)th of full circular cross section.hence perimeter of circular part = (3/4)×2πr = (3/4)×2π×3.5 m ≈ 16.49 m
base = 2×2.47 = 5.94 m
Hence Perimeter of crosssection including base = 16.49+5.94 = 22.43 m
-------------------------------------------------------------
Area of crosssection:
Area of circular portion = (3/4)×π×3.5×3.5
Area of triangular portion = (1/2)×3.5×3.5
total area ≈ 35 sq.m
---------------------------------------------------------
Internal surface area , excluding base, for a length of 50m
surface area = (3/4)×2π×3.5×50 ≈ 825 sq.m
Answered by Thiyagarajan K | 31 Jan, 2019, 10:44: AM
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