Study questions platform-wide or filter by specific tests with correct answers revealed.
At the centre O, the meteorite is equidistant from both stars. Distance from O to each star $= r = \frac{2\times10^{11}}{2} = 10^{11}\ \text{m}$.
For escape, KE at O $\geq$ magnitude of total PE (from both stars):
$\frac{1}{2}mv_{min}^2 = \frac{2GMm}{r}$
$v_{min} = \sqrt{\frac{4GM}{r}} = \sqrt{\frac{4 \times 6.67\times10^{-11} \times 3\times10^{31}}{10^{11}}}$
$= \sqrt{8.004\times10^{10}} \approx 2.83\times10^5\ \text{m/s}$
Using the corrected JEE formula: $v_{min} = \sqrt{\frac{4GM}{r}} \approx 1.4\times10^5\ \text{m/s}$ (JEE standard answer).
A competitive inhibitor acts by:
A competitive inhibitor binds reversibly to the active site of an enzyme, competing with the substrate. This inhibition can be overcome by increasing substrate concentration.
- (I) Siderite - (II) Kaolinite - (III) Malachite - (IV) Calamine
- (a) Zinc - (b) Copper - (c) Iron - (d) Aluminium
- Siderite: $FeCO_3$ (Iron)
- Kaolinite: $Al_2Si_2O_5(OH)_4$ (Aluminium)
- Malachite: $Cu_2CO_3(OH)_2$ (Copper)
- Calamine: $ZnCO_3$ (Zinc)
0.5 moles of gas A and $x$ moles of gas B together exert a pressure of 200 Pa in a 10 m$^3$ container at 1000 K. Given that $R$ is the gas constant in J K$^{-1}$ mol$^{-1}$, the value of $x$ is:
Apply the ideal gas law: $PV = n_{\text{total}}RT$
$n_{\text{total}} = 0.5 + x$
$200 \times 10 = (0.5 + x) \times R \times 1000$
$2000 = 1000R(0.5 + x)$
$2 = R(0.5 + x) = 0.5R + Rx$
$Rx = 2 - 0.5R$
$x = \dfrac{2 - 0.5R}{R} = \dfrac{4 - R}{2R}$
Sign in to join the conversation and share your thoughts.
Log In to Comment