eSaral - JEE | NEET | Class 8-10 Exam Prep App

The cell in which the following reactions occurs:

Question: The cell in which the following reactions occurs: $2 \mathrm{Fe}^{3+}{ }_{(\infty)}+2 \mathrm{I}_{(a q)}^{-} \rightarrow 2 \mathrm{Fe}^{2+}{ }_{(a q)}+\mathrm{I}_{2(s)}$ has $E_{\text {cul }}^{0}=0.236 \mathrm{~V}$ at $298 \mathrm{~K}$. Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction. Solution: Here, $n=2, E_{\text {cell }}^{\ominus}=0.236 \mathrm{~V}, \mathrm{~T}=298 \mathrm{~K}$ We know that: $\Delta_{r} \mathrm{G}^{\ominus}=-n \mathrm{FE}_{\math...

Take the breakdown voltage of the zener diode used in the given circuit as 6V.

Question: Take the breakdown voltage of the zener diode used in the given circuit as 6V. For the input voltage shown in figure below, the time variation of the output voltage is : (Graphs drawn are schematic and not to scale) Correct Option: , 3 Solution: (3) Here two zener diodes are in reverse polarity so if one is in forward bias the other will be in reverse bias and above $6 \mathrm{~V}$ the reverse bias will too be in conduction mode. Hence when $V6 V$ the output will be constant. And when ...

The diameters of the front and rear wheels of a tractor are 80 cm and 2 m respectively.

Question: The diameters of the front and rear wheels of a tractor are 80 cm and 2 m respectively. Find the number of revolutions that rear wheel will make to cover the distance which the front wheel covers in 1400 revolutions. Solution: Let us find the distance covered by front wheel in 1400 revolutions. We know that distance covered in n revolutions is equal to multiplication of number of revolutions and circumference of wheel. $\therefore$ distance $=n \times 2 \times \pi \times r$ We have $r=...

Calculate the emf of the cell in which the following reaction takes place:

Question: Calculate the emf of the cell in which the following reaction takes place: $\mathrm{Ni}_{(s)}+2 \mathrm{Ag}^{+}(0.002 \mathrm{M}) \rightarrow \mathrm{Ni}^{2+}(0.160 \mathrm{M})+2 \mathrm{Ag}_{(s)}$ Given that $E_{\text {(cell) }}^{\ominus}=1.05 \mathrm{~V}$ Solution: ApplyingNernst equation we have: $E_{\text {(cell) }}=E_{\text {(cell) }}^{\ominus}-\frac{0.0591}{n} \log \frac{\left[\mathrm{Ni}^{2+}\right]}{\left[\mathrm{Ag}^{+}\right]^{2}}$ $=1.05-\frac{0.0591}{2} \log \frac{(0.160)}{...

If the points A(x, 2), B(−3, −4) and C(7, −5) are collinear then the value of x is

Question: If the pointsA(x, 2),B(3, 4) andC(7, 5) are collinear then the value ofxis (a) $-63$ (b) 63 (c) 60 (d) $-60$ Solution: LetA(x1=x,y1= 2),B(x2= 3,y2= 4) andC(x3= 7,y3= 5) be collinear points. Then $x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)=0$ $\Rightarrow x(-4+5)+(-3)(-5-2)+7(2+4)=0$ $\Rightarrow x+21+42=0$ $\Rightarrow x=-63$ Hence, the correct answer is option (a)....

Calculate the potential of hydrogen electrode in contact

Question: Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10. Solution: For hydrogen electrode, $\mathrm{H}^{+}+\mathrm{e}^{-} \longrightarrow \frac{1}{2} \mathrm{H}_{2}$, it is given that $\mathrm{pH}=10$ [H+] = 1010M Now, using Nernst equation: = = 0.0591 log 1010 = 0.591 V...

Three circles are placed on a plane in such a way that each circle

Question: Three circles are placed on a plane in such a way that each circle just touches the other two, each having a radius of 10 cm. Find the area of region enclosed by them. Solution: Let there be 3 circles with centre A, B and C respectively such that each circle touches the other two. They all have radius of 10 cm. We have to find the area enclosed between the circles. We have, $\mathrm{AB}=10 \mathrm{~cm}$ $\mathrm{BC}=10 \mathrm{~cm}$ $\mathrm{CA}=10 \mathrm{~cm}$ $A B=10+10 \mathrm{~cm}...

When a diode is forward biased, it has a voltage drop of 0.5 V.

Question: When a diode is forward biased, it has a voltage drop of $0.5 \mathrm{~V}$. The safe limit of current through the diode is 10 $\mathrm{mA}$. If a battery of emf $1.5 \mathrm{~V}$ is used in the circuit, the value of minimum resistance to be connected in series with the diode so that the current does not exceed the safe limit is :$300 \Omega$$50 \Omega$$100 \Omega$$200 \Omega$Correct Option: , 3 Solution: (3) According to question, when diode is forward biased, $V_{\text {diode }}=0.5 \...

If A(1, 3), B(−1, 2) and C(2, 5) and D(x, 4) are the vertices of a ||gm ABCD then the value of x is

Question: IfA(1, 3),B(1, 2) andC(2, 5) andD(x, 4) are the vertices of a ||gmABCDthen the value ofxis (a) 3 (b) 4 (c) 0 (d) $\frac{3}{2}$ Solution: The diagonals of a parallelogram bisect each other. The vertices of the ||gmABCDareA(1, 3),B(1, 2) andC(2, 5) andD(x, 4).Here,ACandBDare the diagonals. So $\frac{1+2}{2}=\frac{-1+x}{2}$ $\Rightarrow x-1=3$ $\Rightarrow x=1+3=4$ Hence, the correct answer is option (b)....

Consult the table of standard electrode potentials

Question: Consult the table of standard electrode potentials and suggest three substances that can oxidise ferrous ions under suitable conditions. Solution: Substances that are stronger oxidising agents than ferrous ions can oxidise ferrous ions. $\mathrm{Fe}^{2+} \longrightarrow \mathrm{Fe}^{3+}+\mathrm{e}^{-1} ; E^{\ominus}=-0.77 \mathrm{~V}$ This impliesthat the substances having higher reduction potentials than+0.77 V can oxidise ferrous ions to ferric ions. Three substances that can do so a...

Can you store copper sulphate solutions in a zinc pot?

Question: Can you store copper sulphate solutions in a zinc pot? Solution: Zinc is more reactive than copper. Therefore, zinc can displace copper from its salt solution. If copper sulphate solution is stored in a zinc pot, then zinc will displace copper from the copper sulphate solution. $\mathrm{Zn}+\mathrm{CuSO}_{4} \longrightarrow \mathrm{ZnSO}_{4}+\mathrm{Cu}$ Hence, copper sulphate solution cannot be stored in a zinc pot....

The perimeter of the triangle with vertices (0, 4), (0, 0) and (3, 0) is

Question: The perimeter of the triangle with vertices (0, 4), (0, 0) and (3, 0) is (a) $7+\sqrt{5}$ (b) 5 (c) 10 (d) 12 Solution: LetA(0, 4),B(0, 0) andC(3, 0) be the given vertices. So $A B=\sqrt{(0-0)^{2}+(4-0)^{2}}=\sqrt{16}=4$ $B C=\sqrt{(0-3)^{2}+(0-0)^{2}}=\sqrt{9}=3$ $A C=\sqrt{(0-3)^{2}+(4-0)^{2}}=\sqrt{9+16}=5$ ThereforeAB+BC+AC= 4 + 3 + 5 = 12Hence, the correct answer is option (d)....

In the following digitial circuit, what will be the output at ' Z ',

Question: In the following digitial circuit, what will be the output at ' $Z$ ', when the input $(A, B)$ are $(1,0),(0,0),(1,1),(0,1)$ : $0,0,1,0$$1,0,1,1$$1,1,0,1$$0,1,0,0$Correct Option: , 3 Solution:...

Find the area of the shaded region in the following figure,

Question: Find the area of the shaded region in the following figure, if AC = 24 cm, BC = 10 cm and O is the centre of the circle. (Use = 3.14) Solution: It is given a triangle ABC is cut from a circle. $A C=24 \mathrm{~cm}$ $B C=10 \mathrm{~cm}$ Area of $\triangle A B C=\frac{1}{2} A C \times B C$ $=\frac{1}{2} \times 24 \times 10$ $=120 \mathrm{~cm}^{2}$ $\ln \triangle A B C$, $\angle A C B=90^{\circ}$, Since any angle inscribed in semicircle is always right angle. By applying Pythagoras theor...

How would you determine the standard electrode potential

Question: How would you determine the standard electrode potential of the systemMg2+| Mg? Solution: The standard electrode potential of Mg2+| Mg can be measured with respect to the standard hydrogen electrode, represented by Pt(s), H2(g)(1 atm) | H+(aq)(1 M). A cell, consisting of Mg | MgSO4(aq1 M) as the anode and the standard hydrogen electrode as the cathode, is set up. $\mathrm{Mg}\left|\mathrm{Mg}^{2+}(\mathrm{aq}, 1 \mathrm{M}) \| \mathrm{H}^{+}(\mathrm{aq}, 1 \mathrm{M})\right| \mathrm{H}...

If the point C(k, 4) divides the join of the points A(2, 6) and B(5, 1) in the ratio 2 : 3 then the value of k

Question: If the pointC(k, 4) divides the join of the pointsA(2, 6) andB(5, 1) in the ratio 2 : 3 then the value ofkis (a) 16 (b) $\frac{28}{5}$ (c) $\frac{16}{5}$ (d) $\frac{8}{5}$ Solution: The pointC(k, 4) divides the join of the pointsA(2, 6) andB(5, 1) in the ratio 2 : 3. So $k=\frac{2 \times 5+3 \times 2}{2+3}=\frac{10+6}{5}=\frac{16}{5}$ Hence, the correct answer is option (c)....

If R(5, 6) is the midpoint of the line segment AB joining the points A(6, 5) and B(4, y) then y equals

Question: IfR(5, 6) is the midpoint of the line segmentABjoining the pointsA(6, 5) andB(4,y) thenyequals (a) 5 (b) 7 (c) 12 (d) 6 Solution: SinceR(5, 6) is the midpoint of the line segmentABjoining the pointsA(6, 5) andB(4,y), therefore $\frac{5+y}{2}=6$ $\Rightarrow 5+y=12$ $\Rightarrow y=12-5=7$ Hence, the correct option is (b)....

The zener diode has a Vs=30 V.

Question: The zener diode has a $V_{s}=30 \mathrm{~V}$. The current passing through the diode for the following ciruit is $\mathrm{mA}$ Solution: $I=\frac{90-30}{4}=15 \mathrm{~mA}$ $\mathrm{I}_{\mathrm{I}}=\frac{30}{5 \mathrm{~K} \Omega}=6 \mathrm{~mA}$ $\mathrm{I}_{2}=15 \mathrm{~mA}-6 \mathrm{~mA}=9 \mathrm{~mA}$...

In the following figure, OABC is a square of side 7 cm.

Question: In the following figure, OABC is a square of side 7 cm. If OAPC is a quadrant of a circle with centre O, then find the area of the shaded region. (Use = 22/7) Solution: Area of shaded region = Area of square OABC Area of quadrant OAPC $=(\text { Side })^{2}-\frac{1}{4} \pi r^{2}$ $=(7)^{2}-\frac{1}{4} \times \frac{22}{7} \times 7 \times 7$ $=49-38.5$ $=10.5 \mathrm{~cm}^{2}$ Hence, the area of the shaded region is $10.5 \mathrm{~cm}^{2}$...

Determine the osmotic pressure of a solution prepared by

Question: Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K2SO4in 2 liter of water at 25 C, assuming that it is completely dissociated. Solution: When $\mathrm{K}_{2} \mathrm{SO}_{4}$ is dissolved in water, $\mathrm{K}^{+}$and $\mathrm{SO}_{4}^{2-}$ ions are produced. $\mathrm{K}_{2} \mathrm{SO}_{4} \longrightarrow 2 \mathrm{~K}^{+}+\mathrm{SO}_{4}^{2-}$ Total number of ions produced = 3 i=3 Given, w= 25 mg = 0.025 g V= 2 L T= 250C = (25 + 273) K = 298 K Also, we kno...

The point on x-axis which is equidistant from points A(−1, 0) and B(5, 0) is

Question: The point onx-axis which is equidistant from pointsA(1, 0) andB(5, 0) is (a) $(0,2)$ (b) $(2,0)$ (c) $(3,0)$ (d) $(0,3)$ Solution: LetP(x, 0) the point onx-axis, then $A P=B P \Rightarrow A P^{2}=B P^{2}$ $\Rightarrow(x+1)^{2}+(0-0)^{2}=(x-5)^{2}+(0-0)^{2}$ $\Rightarrow x^{2}+2 x+1=x^{2}-10 x+25$ $\Rightarrow 12 x=24 \Rightarrow x=2$ Thus, the required point is (2, 0).Hence, the correct answer is option (b)....

From each of the two opposite corners of a square of side 8 cm,

Question: From each of the two opposite corners of a square of side 8 cm, a quadrant of a circle of radius 1.4 cm is cut. Another circle of radius 4.2 cm is also cut from the centre as shown in the following figure. Find the area of the remaining (Shaded) portion of the square. (Use = 22/7) Solution: It is given that a circle of radius 4.2 cm and two quadrants of radius 1.4 cm are cut from a square of side 8 cm. Let the side of square bea.Then, Area of square $=a^{2}$ $=8 \times 8$ $=64 \mathrm{...

Determine the amount of CaCl2 (i = 2.47) dissolved in 2.5 litre

Question: Determine the amount of CaCl2 (i= 2.47) dissolved in 2.5 litre of water such that its osmotic pressure is 0.75 atm at 27C. Solution: We know that, $\pi=i \frac{n}{V} \mathrm{R} T$ $\Rightarrow \pi=i \frac{w}{M V} \mathrm{R} T$ $\Rightarrow w=\frac{\pi M V}{i \mathrm{R} T}$ $\pi=0.75 \mathrm{~atm}$ $V=2.5 \mathrm{~L}$ $i=2.47$ $T=(27+273) \mathrm{K}=300 \mathrm{~K}$ Here, R = 0.0821 L atm K-1mol-1 M = 1 40 + 2 35.5 = 111g mol-1 Therefore, $w=\frac{0.75 \times 111 \times 2.5}{2.47 \times...

The distance of the point (−3, 4) from x-axis is

Question: The distance of the point (3, 4) from x-axis is (a) 3 (b) 3 (c) 4 (d) 5 Solution: The distance of a point $(x, y)$ from $x$-axis is $|y|$. Here, the point is $(-3,4)$. So, its distance from $x$-axis is $|4|=4$. Hence, the correct answer is option (c)....

The circuit contains two diodes each with a forward resistance

Question: The circuit contains two diodes each with a forward resistance of $50 \Omega$ and with infinite reverse resistance. If the battery voltage is $6 \mathrm{~V}$, the current through the $120 \Omega$ resistance is______ $\mathrm{mA}$ Solution: $(\mathbf{2 0})$ $\mathrm{D}_{2}$ is reverse bias so current does not flow through $\mathrm{D}_{2}$ $\mathrm{D}_{1}$ is forward bias. $I=\frac{6}{300} \quad \Rightarrow 0.02 \mathrm{~A}$ $=20 \mathrm{~mA}$...

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