The following is alleged to be a solution to the critical section pr.pdf

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The following is alleged to be a solution to the critical section problem. Argue for its correctness or show a case in which it fails. shared int turn; /* shared variables to synchronize operation */ boolean flag[2]; /* shared variables to synchronize operation */ proc (int i) { while (TRUE) { ; flag[i] = TRUE;/* Attempt to enter the critical section */ turn = (i+1)mod 2; while ((flag[(i+1) mod 2]) && (turn == (i+1) mod 2)) ; /*Now authorized to enter the critical section */ ; /* Exiting the critical section */ flag[i] = FALSE; // Indicate no desire to enter my c.s. } } turn = 0; flag[0] = flag[1] = FALSE; fork (proc, 1, 0); /* Start a process to run proc(0) */ fork (proc, 1, 1); /* Start a process to run proc(1) */ Solution proc (int i) { while (TRUE) { ; flag[i] = TRUE;/* Attempt to enter the critical section */ turn = (i+1)mod 2; while ((flag[(i+1) mod 2]) && (turn == (i+1) mod 2)) ; /*Now authorized to enter the critical section */ ; /* Exiting the critical section */ flag[i] = FALSE; // Indicate no desire to enter my c.s. } } turn = 0; flag[0] = flag[1] = FALSE; fork (proc, 1, 0); /* Start a process to run proc(0) */ fork (proc, 1, 1);.

The following is alleged to be a solution to the critical section problem. Argue for its correctness
or show a case in which it fails.
shared int turn; /* shared variables to synchronize operation */
boolean flag[2]; /* shared variables to synchronize operation */
proc (int i) {
while (TRUE) {
;
flag[i] = TRUE;/* Attempt to enter the critical section */
turn = (i+1)mod 2;
while ((flag[(i+1) mod 2]) && (turn == (i+1) mod 2)) ;
/*Now authorized to enter the critical section */
;
/* Exiting the critical section */
flag[i] = FALSE; // Indicate no desire to enter my c.s.
}
}
turn = 0;
flag[0] = flag[1] = FALSE;
fork (proc, 1, 0); /* Start a process to run proc(0) */
fork (proc, 1, 1); /* Start a process to run proc(1) */
Solution
proc (int i) {
while (TRUE) {
;
flag[i] = TRUE;/* Attempt to enter the critical section */
turn = (i+1)mod 2;
while ((flag[(i+1) mod 2]) && (turn == (i+1) mod 2)) ;
/*Now authorized to enter the critical section */
;
/* Exiting the critical section */
flag[i] = FALSE; // Indicate no desire to enter my c.s.
}

}
turn = 0;
flag[0] = flag[1] = FALSE;
fork (proc, 1, 0); /* Start a process to run proc(0) */
fork (proc, 1, 1);

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Circle the best answer to the following questions. From the Bohr model we learn: The origin of line spectra for glowing hydrogen gas. Those energies for electrons in a hydrogen atom are quantized. The electron orbits have quantized radii All of the above. Silicon is: A rare element in the earth\'s surface A Group V element with five valence electrons. Both A & B. None of the above. The work function of a metal is: The same for different metals. An energy barrier that keeps conduction electrons inside a metal. Has the units of meters. All of the above. In a fictional atom called Hockeygen, the n=3 energy state has an energy of -2.0 eV, and the n=1 energy state has an energy of -5 eV. An electron makes a transition from the n=3 to the n=1 energy state of this atom. We expect: The photon absorbed has an energy of 3 eV. The photon emitted has an energy of 3 eV. The photon emitted has an energy of 5 eV, The photon absorbed has an energy of 5 eV. A typical silicon atom inside a single crystal of silicon has: 4 nearest neighbor dopant atoms. 8 nearest neighbor silicon atoms. 4 nearest neighbor silicon atoms. 4 electrons and 4 holes. For commercial single crystal silicon, typical dopant densities are: 10^15 - 10^17 dopants/cm^3. 10^5-10^7 dopants/cm^3 5 times 10^22 dopants/cm^3 6.02 times 10^23 dopants/cm^3 The type of solid that has the highest degree of atomic order is: Single crystal. Polycrystalline. Amorphous. Metals. Solution Problem 1 a :The model\'s key success lay in explaining the Rydberg formula for the spectral emission lines of atomic hydrogen. While the Rydberg formula had been known experimentally, it did not gain a theoretical underpinning until the Bohr model was introduced. Not only did the Bohr model explain the reason for the structure of the Rydberg formula, it also provided a justification for its empirical results in terms of fundamental physical constants. Answer: 1 Problem 1 b : Measured by mass, silicon makes up 27.7% of the Earth\'s crust and is the second most abundant element in the crust, with only oxygen having a greater abundance.Silicon is usually found in the form of complex silicate minerals, and less often assilicon dioxide (silica, a major component of common sand). It belongs to the 14th group of periodic table. Answer 4 Problem 1 c : the minimum quantity of energy which is required to remove an electron to infinity from the surface of a given solid, usually a metal. Answer 2 Problem 1 d: ENERGY required to make transition is equal to E1-E3 = -5-(-2) = -3eV So a photon with energy = 3eV is released in transition. Answer 2 Problem 1 e: For single-crystal silicon, each silicon atom has four nearest neighbors. Since Si is a column IV element and has four valence electrons, each of these electrons is shared with one of its four nearest neighbor Si atoms. Answer 3 Problem 1 f: In intrinsic crystalline silicon, there are approximately 5×1022 atoms/cm³. Doping concentration for silicon semiconductors may range anywhere .

Circle the best answer to the following questions. From the Bohr mod.pdf

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Below are two possible tree topologies. How many clades are different, in terms of their associated tips? Define exaptation. Give an example of an exaptation. Define polyphenism and give an example. Draw an illustrative isochron plot based upon 87Sr/86Sr and 87Rb/86Sr. Draw lines with different slopes, and indicate which slope corresponds to the more distant moment in time? Solution Q5. Answer: The topologies in the trees are different considering the the first tree it has five clades The B and F have major similllarity hence they have made a clade which is considered as not much variation in the genotype. next comes A and C which has made the clade as they have less variation rest D and E have more variation with other members of the clade as they form the solitary clade. Comparing first tree with the second thay have common clade A and C where as DB and F had varied lot with respect to clad formation but Clade E has formed solitary clade in both the topological trees. Q6. Answer : Exaptation is a term used in evolutionary biology to describe a trait that has been co-opted for a use other than the one for which natural selection has built it. The feathers in birds which are intially used for the heat regulation of the body, which were co- opted for dispaly and flight lateron . A behavioural example in wolves, licking the mouths of alpha wolves as a sign of submissiveness. (Similarly, dogs, which are wolves who through a long process were domesticated, lick the faces of their human owners.) This trait can be explained as an exaptation of wolf pups licking the faces of adults to encourage them to regurgitate food. Q7 .A biological mechanism that causes a trait to be polyphenic is called polyphenism. For example, crocodiles possess a sex-determining polyphenism, and therefore their gender is a polyphenic trait. Polyphenic pigmentation is adaptive for insect species that undergo multiple mating seasons each year. Different pigmentation patterns provide appropriate camouflage throughout the seasons, as well as alter heat retention as temperatures change. Because insects cease growth and development after eclosion, their pigment pattern is invariable in adulthood: thus, a polyphenic pigment adaptation would be less valuable for species whose adult form survives longer than one year.

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