A) \[ab\]
B) 1
C) \[bc\]
D) 0
Correct Answer: B
Solution :
The three planes can be written as \[\cos \,\theta =\frac{\sqrt{2}v}{2v}=\frac{1}{\sqrt{2}}=\cos \,{{45}^{o}}\] ...(i) \[\therefore \] ...(ii) \[\times \] ...(iii) Let l,m and n be the DC's of the line of intersection of Eqs. (i) and (ii), then \[=(mg\,\sin \,\theta )\,\frac{a}{2}\] \[\frac{V}{t}=\frac{\pi p{{r}^{4}}}{8\eta l}\] Now, by cross multiplication method, \[{{V}_{1}}=\frac{p{{r}^{4}}}{8\eta \left( l+\frac{l}{2} \right)}=\frac{2}{3}\times \frac{\pi p{{r}^{4}}}{8\eta l}\] Since, planes (i) and (ii) passes through origin. Hence, there line of intersection will also passes through the point (0, 0, 0). So, the equation of the line of intersection of Eqs. (i) and (ii) is \[[\because \,\,{{l}_{1}}=l=2{{l}_{2}}\,\,or\,\,{{l}_{2}}=l/2]\] ?(iv) Now, the three planes will intersect in a line of Eq. (iv) satisfies Eq. (iii). Hence, the required condition is \[\therefore \] \[{{V}_{1}}=\frac{2}{3}\times 8=\frac{16}{3}\,c{{m}^{3}}/s\] \[=E\times g\pi {{R}^{2}}=\pi {{R}^{2}}E\]
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