Find the cost of sinking a tube-well 280 m deep, having a diameter 3 m at the rate of ₹ 15 per cubic metre.
Question: Find the cost of sinking a tube-well 280 m deep, having a diameter 3 m at the rate of ₹ 15 per cubic metre. Find also the cost of cementing its inner curved surface at ₹ 10 per square metre. Solution: Height of the tube-well,h = 280 m Radius of the tube-well, $r=\frac{3}{2} \mathrm{~m}$ $\therefore$ Volume of the tube-well $=\pi r^{2} h=\frac{22}{7} \times\left(\frac{3}{2}\right)^{2} \times 280=1980 \mathrm{~m}^{3}$ Rate of sinking the tube-well =₹ 15 per cubic metre Cost ofsinking the...
Read More →The major product obtained in the following reaction is:
Question: The major product obtained in the following reaction is: Correct Option: , 4 Solution: Reaction involved:...
Read More →A cylindrical container with diameter of base 56 cm contains sufficient water to submerge a rectangular solid of iron with dimensions (32 cm × 22 cm × 14 cm).
Question: A cylindrical container with diameter of base 56 cm contains sufficient water to submerge a rectangular solid of iron with dimensions (32 cm 22 cm 14 cm). Find the rise in the level of water when the solid is completely submerged. Solution: Volume of the rectangular solid of iron = 32 cm 22 cm 14 cm Radius of the container, $r=\frac{56}{2}=28 \mathrm{~cm}$ Let the rise in the level of water in the container when rectangular solid of iron is submerged in itbehcm. $\therefore$ Volume of ...
Read More →A particle of mass m moves in a circular orbit in a central potential field
Question: A particle of mass $\mathrm{m}$ moves in a circular orbit in a central potential field $\mathrm{U}(\mathrm{r})=\mathrm{U}_{0} r^{4}$. If Bohr's quantization conditions are applied, radii of possible orbitals $\mathrm{r}_{\mathrm{n}}$ vary with $\mathrm{n}^{1 / \alpha}$, where $\alpha$ is Solution: (3) $\mathrm{F}=\frac{-\mathrm{dU}}{\mathrm{dr}}=-4 \mathrm{U}_{0} \mathrm{r}^{3}=\frac{\mathrm{mv}^{2}}{\mathrm{r}}$ $\mathrm{mv}^{2}=4 \mathrm{U}_{0} \mathrm{r}^{4}$ $\mathrm{V} \propto \ma...
Read More →A cylindrical water tank of diameter 1.4 m and height 2.1 m is being fed by a pipe of diameter 3.5 cm through which water flows at the rate of 2 m per second.
Question: A cylindrical water tank of diameter 1.4 m and height 2.1 m is being fed by a pipe of diameter 3.5 cm through which water flows at the rate of 2 m per second. In how much time will the tank be filled? Solution: Radius of the water $\tan k, R=\frac{1.4}{2}=0.7 \mathrm{~m}$ Height of the water tank,H= 2.1 m $\therefore$ Capacity of the water tank $=\pi R^{2} H=\pi(0.7)^{2} \times 2.1 \mathrm{~m}^{3}$ Speed of the water flow = 2 m/s Radius of the pipe, $r=\frac{3.5}{2}=1.75 \mathrm{~cm}=0...
Read More →The major product of the following reaction is:
Question: The major product of the following reaction is: Correct Option: , 4 Solution: Reaction involved:...
Read More →How many litres of water flows out of a pipe having an area of cross section of 5 cm2 in 1 minute,
Question: How many litres of water flows out of a pipe having an area of cross section of 5 cm2in 1 minute, if the speed of water in the pipe is 30 cm/sec? Solution: Speed of the water = 30 cm/sArea of the cross section = 5 cm2Volume of the water flowing out of the pipe in one second = Area of the cross section 30 cm = 5 30 = 150 cm3Now,1 minute = 60 secondsVolume of the water flowing out of the pipe in 60 seconds= Volume of the water flowing out of the pipe in one second60=15060= 9000 cm3 $=\fr...
Read More →A well with inside diameter 10 m is dug 8.4 m deep. Earth taken out of it is spread all around it to a width of 7.5 m to form an embankment.
Question: A well with inside diameter 10 m is dug 8.4 m deep. Earth taken out of it is spread all around it to a width of 7.5 m to form an embankment. Find the height of the embankment. Solution: Inner radius of the well, $r=\frac{10}{2}=5 \mathrm{~m}$ Depth of the well,h= 8.4 mSuppose the outer radius of the embankment isRm.Width of the embankment = 7.5 mRr= 7.5 m⇒R= 7.5 + 5 = 12.5 mLet the height of the embankment beHm.Now,Volume of earth used to form the embankment = Volume of earth dugged ou...
Read More →Two identical photocathodes receive the light
Question: Two identical photocathodes receive the light of frequencies $f_{1}$ and $f_{2}$ respectively. If the velocities of the photo-electrons coming out are $\mathrm{v}_{1}$ and $\mathrm{v}_{2}$ respectively, then(1) $\mathrm{v}_{1}^{2}-\mathrm{v}_{2}^{2}=\frac{2 \mathrm{~h}}{\mathrm{~m}}\left[\mathrm{f}_{1}-\mathrm{f}_{2}\right]$(2) $\mathrm{v}_{1}^{2}+\mathrm{v}_{2}^{2}=\frac{2 \mathrm{~h}}{\mathrm{~m}}\left[\mathrm{f}_{1}+\mathrm{f}_{2}\right]$(3) $\mathrm{v}_{1}+\mathrm{v}_{2}=\left[\fra...
Read More →Solve the following
Question: Correct Option: 4, Solution: $\mathrm{NaBH}_{4}$ is a selective reducing agent, used for the reduction of aldehydes and ketones, it does not affect alkene and ester....
Read More →A helicopter is flying along the curve given by
Question: A helicopter is flying along the curve given by $y-x^{3 / 2}=7,(x \geq 0)$. A soldier positioned at the point $\left(\frac{1}{2}, 7\right)$ wants to shoot down the helicopter when it is nearest to him. Then this nearest distance is: (1) $\frac{\sqrt{5}}{6}$(2) $\frac{1}{3} \sqrt{\frac{7}{3}}$(3) $\frac{1}{6} \sqrt{\frac{7}{3}}$(4) $\frac{1}{2}$Correct Option: , 3 Solution: $f(x)=y=x^{3 / 2}+7$ $\Rightarrow \quad \frac{d y}{d x} \Rightarrow \frac{3}{2} \sqrt{x}0$ $\Rightarrow f(x)$ is i...
Read More →A juiceseller has a large cylindrical vessel of base radius 15 cm filled up to a height of 32 cm with orange juice.
Question: A juiceseller has a large cylindrical vessel of base radius 15 cm filled up to a height of 32 cm with orange juice. The juice is filled in small cylindrical glasses of radius 3 cm up to a height of 8 cm, and sold for ₹ 15 each. How much money does he received by selling the juice completely? Solution: Radius of the vassel,R= 15 cmHeight of orange juice in the vassel,H= 32 cm $\therefore$ Volume of orange juice in the vassel $=\pi R^{2} H=\pi \times(15)^{2} \times 32 \mathrm{~cm}^{3}$ R...
Read More →The tests performed on compound X and their inferences are:
Question: The tests performed on compound $\mathrm{X}$ and their inferences are: Correct Option: 2, Solution: Reaction involved in the tests performed is as follows:...
Read More →It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet.
Question: It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. How many square metres of the sheet are required for the same? Solution: Height of the cylindricaltank,h= 1 m Radius of the cylindrical tank, $r=\frac{140}{2}=70 \mathrm{~cm}=0.7 \mathrm{~m} \quad(1 \mathrm{~m}=100 \mathrm{~cm})$ Area of the metal sheet required to make the cylindricaltank= Total surface area of the cylindrical tank $=2 \pi r h(r+h)$ $=2 \times \frac{22}{7} \time...
Read More →The tests performed on compound $mathrm{X}$ and their inferences are:
Question: The tests performed on compound $\mathrm{X}$ and their inferences are: Correct Option: Solution: Reaction involved in the tests performed is as follows:...
Read More →The tangent to the curve,
Question: The tangent to the curve, $y=x \mathrm{e}^{x^{2}}$ passing through the point $(1$, e $)$ also passes through the point:(1) $(2,3 \mathrm{e})$(2) $\left(\frac{4}{3}, 2 \mathrm{e}\right)$(3) $\left(\frac{5}{3}, 2 \mathrm{e}\right)$(4) $(3,6 \mathrm{e})$Correct Option: , 2 Solution: The equation of curve $y=x e^{x^{2}}$ $\Rightarrow \quad \frac{d y}{d x}=e^{x^{2}} \cdot 1+x, \mathrm{e}^{x^{2}} \cdot 2 x$ Since $(1, e)$ lies on the curve $y=x e^{x^{2}}$, then equation of tangent at $(1, e)...
Read More →The atomic hydrogen emits a line spectrum consisting of various series.
Question: The atomic hydrogen emits a line spectrum consisting of various series. Which series of hydrogen atomic spectra is lying in the visible region?(1) Brackett series(2) Paschen series(3) Lyman series(4) Balmer seriesCorrect Option: Solution: (4) Conceptual...
Read More →A cylindrical tube, open at both ends, is made of metal.
Question: A cylindrical tube, open at both ends, is made of metal. The internal diameter of the tube is 10.4 cm and its length is 25 cm. The thickness of the metal is 8 mm everywhere. Calculate the volume of the metal. Solution: Inner radius of the cylindrical tube,r= 5.2 cmHeight of the cylindrical tube,h= 25 cmOuter radius of the cylindrical tube,R =(5.2 + 0.8) cm = 6 cm $\therefore$ Volume of the metal $=$ external volume $-$ internal volume $=\pi R^{2} h-\pi r^{2} h \quad$ (where $R$ and $r$...
Read More →If I is minimum then the ordered pair
Question: Let $\mathrm{I}=\int_{\mathrm{a}}^{\mathrm{b}}\left(x^{4}-2 x^{2}\right) \mathrm{d} x$. If I is minimum then the ordered pair $(a, b)$ is:(1) $(0, \sqrt{2})$(2) $(-\sqrt{2}, 0)$(3) $(\sqrt{2},-\sqrt{2})$(4) $(-\sqrt{2}, \sqrt{2})$Correct Option: , 4 Solution: $I=\int_{a}^{b}\left(x^{4}-2 x^{2}\right) d x$ $\Rightarrow \frac{d I}{d x}=x^{4}-2 x^{2}=0$ (for minimum) $\Rightarrow x=0, \pm \sqrt{2}$ Also$I=\left[\frac{x^{5}}{5}-\frac{2 x^{3}}{3}\right]_{a}^{b}$ For $\mathrm{a}=0, \mathrm{~...
Read More →If 1 cm3 of case iron weighs 21 g, find the weight of a cast iron pipe of length
Question: If 1 cm3of case iron weighs 21 g, find the weight of a cast iron pipe of length 1 m with a bore of 3 cm in which the thickness of the metal is 1 cm. Solution: Internal radius of the pipe = 1.5 cmExternal radius of the pipe = (1.5 + 1) cm = 2.5 cmHeight of the pipe = 1 m = 100 cmVolume of the cast iron = total volume of the pipe internal volume of the pipe $=\pi \times(2.5)^{2} \times 100-\pi \times(1.5)^{2} \times 100$ $=\frac{22}{7} \times 100 \times[6.25-2.25]$ $=\frac{2200}{7} \time...
Read More →Which level of the single ionized carbon has
Question: Which level of the single ionized carbon has the same energy as the ground state energy of hydrogen atom?(1) 1(2) 6(3) 4(4) 8Correct Option: , 2 Solution: (2) Energy of $\mathrm{H}$-atom is $\mathrm{E}=-13.6 \mathrm{Z}^{2} / \mathrm{n}^{2}$ for $\mathrm{H}$-atom $\mathrm{Z}=1 \$ for ground state, $\mathrm{n}=1$ $\Rightarrow \mathrm{E}=-13.6 \times \frac{1^{2}}{1^{2}}=-13.6 \mathrm{eV}$ Now for carbon atom (single ionised), $Z=6$ $\mathrm{E}=-13.6 \frac{\mathrm{Z}^{2}}{\mathrm{n}^{2}}=-...
Read More →1 cm3 of gold is drawn into a wire 0.1 mm in diameter.
Question: 1 cm3of gold is drawn into a wire 0.1 mm in diameter. Find the length of the wire. Solution: Letrcm be the radius of the wire andhcm be the height of the wire.Volume of the gold = 1 cm31 cm3of gold is drawn into a wire of diameter 0.1 mm. Here, $r=0.1 \mathrm{~mm}=\frac{0.1}{20} \mathrm{~cm}=\frac{1}{200} \mathrm{~cm}$ $\therefore \pi r^{2} h=1$ $\Rightarrow \frac{22}{7} \times \frac{1}{200} \times \frac{1}{200} \times h=1$ $\Rightarrow h=\frac{40000 \times 7}{22}=12727.27 \mathrm{~cm}...
Read More →The major product formed in the following reaction is:
Question: The major product formed in the following reaction is: Correct Option: Solution: Reaction mechanism involved:...
Read More →The barrel of a fountain pen, cylindrical in shape, is 7 cm long and 5 mm in diameter.
Question: The barrel of a fountain pen, cylindrical in shape, is 7 cm long and 5 mm in diameter. A full barrel of ink in the pen will be used up on writing 330 words on an average. How many words would use up a bottle of ink containing one-fifth of a litre? Solution: Height of the barrel =h= 7 cmRadius of the barrel =r= 2.5 mm = 0.25 cm Volume of the barrel $=\pi r^{2} h$ $=\frac{22}{7} \times 0.25 \times 0.25 \times 7$ $=22 \times 0.25 \times 0.25$ $=1.375 \mathrm{~cm}^{3}$ i.e., 1.375 cm3of in...
Read More →The major product of following reaction is:
Question: The major product of following reaction is: $\mathrm{RCOOH}$$\mathrm{RCONH}_{2}$$\mathrm{RCHO}$$\mathrm{RCH}_{2} \mathrm{NH}_{2}$Correct Option: , 3 Solution: The reduction of nitriles to aldehydes can be done using DIBAL-H[AIH(i-Bu) $_{2}$ ]....
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