The question asks about the direction of currents in the arms of a coil in an electric motor, specifically in relation to Fleming's Left-Hand Rule. This rule helps determine the direction of force, magnetic field, and current. In an electric motor, the coil rotates due to forces acting on its arms. These forces are generated by the interaction of the current flowing through the coil and the magnetic field.
Correct Option: A) opposite. The currents in the two active arms of the coil (the arms parallel to the axis of rotation) must flow in opposite directions to produce forces that create a turning effect (torque) on the coil, causing it to rotate. This is a fundamental requirement for the operation of an electric motor, as dictated by Fleming's Left-Hand Rule.
To find the quadratic equation given its roots, we can use the relationship between the roots and the coefficients of a quadratic equation. A quadratic equation with roots \(\alpha\) and \(\beta\) can be expressed as \(x^2 - (\alpha + \beta)x + \alpha\beta = 0\).
The roots of the quadratic equation are given as \(\alpha = 2\) and \(\beta = -3\).
The sum of the roots is \(\alpha + \beta\).
\[ \text{Sum} = 2 + (-3) = 2 - 3 = -1 \]The product of the roots is \(\alpha\beta\).
\[ \text{Product} = 2 \times (-3) = -6 \]The general form of a quadratic equation with roots \(\alpha\) and \(\beta\) is:
\[ x^2 - (\text{Sum of roots})x + (\text{Product of roots}) = 0 \]Substitute the sum of roots (\(-1\)) and the product of roots (\(-6\)) into the general form:
\[ x^2 - (-1)x + (-6) = 0 \]\[ x^2 + x - 6 = 0 \]Correct Option: C) x² + x - 6 = 0
This equation would have a product of roots equal to \(+6\), but our product of roots is \(-6\).
This equation would have a sum of roots equal to \(+1\), but our sum of roots is \(-1\).
This equation would have a sum of roots equal to \(+1\) and a product of roots equal to \(+6\), both of which are incorrect.
The question asks to identify the false statement about HIV. We need to evaluate each option based on known facts about HIV and its mechanism of action in the human body.
Based on this analysis, statement C is false.
Correct Option: C) HIV enters your red blood cells and attacks your immune system.
The question asks to identify the type of forest that provides resources like silver fir, juniper, pine, and birch, which are typically found in mountainous regions and used by local tribes and for shrubs and herbs. We need to consider the characteristics and geographical distribution of each forest type mentioned in the options.
C) Mountain Forest. Mountain forests are characterized by a succession of vegetation types depending on altitude. At higher altitudes, coniferous trees like pine, fir, spruce, and juniper thrive. Birch is also found in these regions, often at moderate to high elevations. These forests are the primary source of the mentioned resources for communities and ecosystems in mountainous terrains.
To find the effective focal length and power of a doublet made of two thin lenses in contact, we use the formula for combined focal length and the definition of optical power. For lenses in contact, the reciprocal of the effective focal length is the sum of the reciprocals of individual focal lengths. The power of a lens is the reciprocal of its focal length in meters.
For two thin lenses in contact, the formula for the effective focal length is:
\[ \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} \]Substitute the given values:
\[ \frac{1}{F} = \frac{1}{+20} + \frac{1}{-50} \] \[ \frac{1}{F} = \frac{1}{20} - \frac{1}{50} \]Find a common denominator (100):
\[ \frac{1}{F} = \frac{5}{100} - \frac{2}{100} \] \[ \frac{1}{F} = \frac{3}{100} \]Therefore, the effective focal length is:
\[ F = \frac{100}{3} \text{ cm} \approx 33.33 \text{ cm} \]The power of a lens is given by \(P = \frac{1}{F}\), where \(F\) is in meters. First, convert the effective focal length to meters:
\[ F = \frac{100}{3} \text{ cm} = \frac{100}{3 \times 100} \text{ m} = \frac{1}{3} \text{ m} \]Now, calculate the power:
\[ P = \frac{1}{F} = \frac{1}{(1/3) \text{ m}} = 3 \text{ Diopters (D)} \]Correct Option: B) 33.3 cm, 3D
The question asks about the phenomenon that occurs when a beam of fine white light passes through a prism. This involves understanding the interaction of light with a prism, specifically how different colors within white light behave.
D) The ray will be dispersed — When white light passes through a prism, it splits into its constituent colors (VIBGYOR). This phenomenon is called dispersion, which occurs because the refractive index of the prism material is different for different wavelengths of light, causing each color to bend at a slightly different angle.
Consider a circuit in which three resistors of 10 Ω, 20 Ω and 30 Ω are connected in parallel to each other to a cell of 220 V. The effective resistance and the current in the 20 Ω resistor are ______ and ______, respectively.
\( \frac{30}{11} \) Ω, 22 A
\( \frac{60}{11} \) Ω, 11 A
\( \frac{60}{11} \) Ω, 22 A
\( \frac{30}{11} \) Ω, 11 A
To solve this problem, we need to calculate two quantities: the effective resistance of three resistors connected in parallel and the current flowing through one of the parallel resistors. We will use the formula for equivalent resistance in a parallel circuit and Ohm's Law.
Calculate the effective resistance (\(R_{eq}\)) of the parallel circuit:
For resistors connected in parallel, the reciprocal of the equivalent resistance is the sum of the reciprocals of individual resistances. The given resistances are \(R_1 = 10 \, \Omega\), \(R_2 = 20 \, \Omega\), and \(R_3 = 30 \, \Omega\).
\[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \] \[ \frac{1}{R_{eq}} = \frac{1}{10 \, \Omega} + \frac{1}{20 \, \Omega} + \frac{1}{30 \, \Omega} \]Find a common denominator, which is 60:
\[ \frac{1}{R_{eq}} = \frac{6}{60 \, \Omega} + \frac{3}{60 \, \Omega} + \frac{2}{60 \, \Omega} \] \[ \frac{1}{R_{eq}} = \frac{6 + 3 + 2}{60 \, \Omega} = \frac{11}{60 \, \Omega} \]Therefore, the effective resistance is:
\[ R_{eq} = \frac{60}{11} \, \Omega \]Calculate the current in the \(20 \, \Omega\) resistor:
In a parallel circuit, the voltage across each resistor is the same as the voltage of the source. The cell voltage is \(V = 220 \, \text{V}\).
Using Ohm's Law, \(I = \frac{V}{R}\), for the \(20 \, \Omega\) resistor:
\[ I_{20\Omega} = \frac{V}{R_2} \] \[ I_{20\Omega} = \frac{220 \, \text{V}}{20 \, \Omega} \] \[ I_{20\Omega} = 11 \, \text{A} \]Thus, the effective resistance is \( \frac{60}{11} \, \Omega \) and the current in the \(20 \, \Omega\) resistor is \( 11 \, \text{A} \).
B) \( \frac{60}{11} \) Ω, 11 A. This option matches both calculated values: the effective resistance and the current through the \(20 \, \Omega\) resistor.
The length of each side of a rhombus is 20 cm, and the length of one of its diagonals is 30 cm. Find the length (in cm) of its other diagonal.
\(\sqrt{175}\)
10\(\sqrt{7}\)
\(\sqrt{7}\)
6\(\sqrt{7}\)
The problem involves finding the length of a diagonal of a rhombus given the length of its side and one diagonal. We will use the properties of a rhombus, specifically that its diagonals bisect each other at right angles. This property allows us to form right-angled triangles, to which we can apply the Pythagorean theorem.
Let the rhombus be ABCD, with side length \(a = 20 \text{ cm}\). Let the diagonals be \(d_1\) and \(d_2\). Let \(d_1 = 30 \text{ cm}\).
The diagonals of a rhombus bisect each other at right angles. Let the point of intersection of the diagonals be O.
Since the diagonals bisect each other, the half-lengths of the diagonals are \(\frac{d_1}{2}\) and \(\frac{d_2}{2}\).
Given \(d_1 = 30 \text{ cm}\), its half-length is \[\frac{d_1}{2} = \frac{30}{2} = 15 \text{ cm}\].
Consider one of the right-angled triangles formed by a side of the rhombus and the half-lengths of the diagonals, for example, triangle AOB. In \(\triangle AOB\), the hypotenuse is the side of the rhombus (\(AB = a = 20 \text{ cm}\)), and the legs are \(\frac{d_1}{2}\) and \(\frac{d_2}{2}\).
Applying the Pythagorean theorem to \(\triangle AOB\):
\[ \left(\frac{d_1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2 = a^2 \]Substitute the known values:
\[ (15)^2 + \left(\frac{d_2}{2}\right)^2 = (20)^2 \]Calculate the squares:
\[ 225 + \left(\frac{d_2}{2}\right)^2 = 400 \]Isolate the term with \(d_2\):
\[ \left(\frac{d_2}{2}\right)^2 = 400 - 225 \]\[ \left(\frac{d_2}{2}\right)^2 = 175 \]Take the square root of both sides:
\[ \frac{d_2}{2} = \sqrt{175} \]Simplify \(\sqrt{175}\). We know that \(175 = 25 \times 7\):
\[ \frac{d_2}{2} = \sqrt{25 \times 7} \]\[ \frac{d_2}{2} = 5\sqrt{7} \]Solve for \(d_2\):
\[ d_2 = 2 \times 5\sqrt{7} \]\[ d_2 = 10\sqrt{7} \text{ cm} \]B) \(10\sqrt{7}\)
The length of the other diagonal is \(10\sqrt{7}\) cm, which matches option B.
The question asks to identify which environmental effect is NOT caused by the burning of fossil fuels. We need to evaluate each option based on the known impacts of fossil fuel combustion.
Correct Option: D) Decrease in ozone level in the upper atmosphere