The correct answer is: $ \displaystyle 3\sqrt{3}$ cm Explanation: Let P be an external point and a pair of tangents is drawn from point P and angle between these two tangents is 60°. Join OA and OP. Also, OP is a bisector line of ∠APC. ∴ ​∠APO = ∠CPO = 30°  Also,  OA⊥AP​ Tangent at any […]

The correct answer is: $ \displaystyle 3\sqrt{3}$ cm Explanation: Let P be an external point and a pair of tangents is drawn from point P and angle between these two tangents is 60°. Join OA and OP. Also, OP is a bisector line of ∠APC. ∴ ​∠APO = ∠CPO = 30°  Also,  OA⊥AP​ Tangent at any

The correct answer is: 693 cm² Explanation: Since the ribs are equally spaced, so the angle made by two consecutive ribs at the centre =($\displaystyle \dfrac{360}{8}$​)° = 45° Area between two consecutive ribs = area of a sector of a circle with r = 42 cm and θ = 45° =($\displaystyle \dfrac{45}{3600}$ × $\displaystyle \dfrac{22}{7}$ ​× 42

The correct answer is: 9.625 cm² Explanation: Area of shaded region = $\displaystyle \dfrac{22×30°}{7×360°}$  × (7² − 3.5²) = 9.625cm²

$ \displaystyle \dfrac{120o}{2}$

The correct answer is: $ \displaystyle \text{84 c}{{\text{m}}^{2}}$ Explanation: The semi-perimeter $ \displaystyle s=\dfrac{{13+14+15}}{2}=21\text{ cm}$ Using Heron’s formula: Area $ \displaystyle \begin{array}{l}=\sqrt{{s\left( {s-a} \right)\left( {s-b} \right)\left( {s-c} \right)}}=\sqrt{{21\left( {21-13} \right)\left( {21-14} \right)\left( {21-15} \right)}}\\=\sqrt{{21\times 8\times 7\times 6}}=\sqrt{{7056}}=84\text{ c}{{\text{m}}^{2}}\end{array}$

4 kW Explanation: Total power output = Area × Solar irradiance × Efficiency = 25 m² × 800 W/m² × 0.20 = 4,000 W = 4 kW

The correct answer is: 17: 21 Explanation: We known that ratio of areas of two similar triangles is equal to the ratio of square of their corresponding heights. $\displaystyle \dfrac{ar(△ABC)}{ar . (△DEF)}$ = $\displaystyle \dfrac{(4)²}{(9)²}$ =  $\displaystyle \dfrac{16}{81}$ ​

$latex \displaystyle \begin{array}{l}(\text{Take g}=9.8m/{{s}^{2}})\\{{v}_{y}}=v\sin \theta \\so,80\sin {{60}^{\circ }}=80\times 0.866=69.28m/s\\\text{Maximum height }h=/2g={{(69.28)}^{2}}/2\times 9.8\approx 244.9m\end{array}$

$\displaystyle \begin{array}{l}\text{Resultant velocity i}\text{.e}\text{.}\\{{v}_{{re}}}=\sqrt{{{{v}_{s}}^{2}+{{v}_{r}}^{2}}}\\\text{So, }\sqrt{{{{5}^{2}}+{{3}^{2}}}}=\sqrt{{25+9}}=\sqrt{{34}}\\\approx 5.83~m/s\end{array}$